The discriminant
Why does a quadratic equation sometimes have two solutions, sometimes one, and sometimes none at all, and how can you tell which before solving it?
Where you are 4 in The quadratic formula
Why this exists
Solve three quadratics that look almost the same and you can get three different kinds of answer. x^2 - 4x + 3 = 0 has two solutions, 1 and 3. x^2 - 4x + 4 = 0 has one, 2. x^2 - 4x + 5 = 0 has none among the real numbers. Nothing in the look of the equations warns you which is which.
Often the count is all you want. Does a thrown ball ever reach a given height? Are there two numbers with a given sum and product? Does a line meet a curve? Each asks whether a quadratic has a solution, not what it is, and solving in full is a lot of work to learn one word.
The quadratic formula already holds the answer. Split its fraction in two:
The first piece, \displaystyle -\frac{b}{2a}, is the same for both roots: it is the middle, the point the two roots sit either side of. Everything that makes the two roots different from each other lives in the second piece, and inside it the only thing that can change the story is the number under the square root. Whether the roots are whole numbers, fractions, square roots or not real at all; how far apart they are; whether there are two of them or they merge into one; where the parabola meets the x-axis: all of it is settled by b^2 - 4ac. A number that settles that much deserves a name of its own.
Definition
What the discriminant is
Take a quadratic equation in standard form,
with a, b and c real numbers. Its discriminant is the number under the square root in the quadratic formula,
In words: square the coefficient of x, then take away four times the product of the coefficient of x^2 and the constant term. Many books write \Delta for it instead of D. The name comes from the Latin for “to tell apart”. Three details make it the right number rather than a nearby wrong one.
Standard form first. a, b and c are read off only once every term is on one side and 0 is on the other. In 3x^2 = 2x + 5 the value of b is not 2; after moving everything across, the equation is 3x^2 - 2x - 5 = 0 and b = -2.
Each coefficient keeps its sign. In x^2 - 4x + 7 = 0, b = -4, so b^2 = (-4)^2 = 16. A term that is missing has coefficient 0.
It is one number. The discriminant has no x in it. It is not a solution; it is a fact about the equation, computed from its three coefficients.
Roots, and how the graph meets the axis
A root of the equation is a number that makes it true, which is the same thing as a solution. A real root is one on the number line. When this lesson counts roots, it counts different real numbers: an equation whose only solution is 2 has one real root, even though 2 comes out of the formula twice.
The real roots of ax^2 + bx + c = 0 are the x-intercepts of the parabola y = ax^2 + bx + c, and the discriminant of that parabola means the discriminant of that equation. The graph and the x-axis can meet in three ways. The parabola crosses the axis when it passes from one side to the other, which a parabola does at two points or not at all. It touches the axis when it meets it at one point and turns back without crossing, which can only happen at the vertex. And it misses the axis when the whole curve lies on one side.
The claim of this lesson, in plain words: the sign of the discriminant tells you how many real roots there are. Positive means two, and the graph crosses the axis twice. Zero means one, and the graph touches. Negative means none that are real, the graph misses, and the formula gives two complex roots instead.
Intuition
Every quadratic is a square equal to D
Multiply ax^2 + bx + c = 0 by 4a and tidy up (the steps are under Why it is true below), and every quadratic equation becomes a single line:
So solving a quadratic is really one question in disguise: how many numbers square to D? Call such a number u. The picture of y = u^2 answers it. A level above the axis meets the curve twice, at \displaystyle u = \sqrt{D} and \displaystyle u = -\sqrt{D}; the level 0 meets it only at the bottom; a level below the axis never meets it, because no real square is negative.
Getting back from u to x loses nothing and adds nothing. The rule u = 2ax + b is a straight line with slope 2a \ne 0, so each u comes from exactly one x, and different u come from different x. Two square roots give two roots, one gives one, none gives none.
Try it on the equations above. For x^2 - 4x + 3 = 0, D = 16 - 12 = 4, and the equation is (2x - 4)^2 = 4. So 2x - 4 = 2 or 2x - 4 = -2, giving x = 3 or x = 1. For x^2 - 4x + 4 = 0, D = 0, so (2x - 4)^2 = 0 and x = 2 only. For x^2 - 4x + 5 = 0, D = -4, and (2x - 4)^2 = -4 asks for a real square that is negative. There is none.
The middle and the gap
Take x^2 - 6x + 5 = 0, so a = 1, b = -6 and c = 5. Then
The middle is \displaystyle -\frac{b}{2a} = \frac{6}{2} = 3, and each root sits \displaystyle \frac{\sqrt{16}}{2} = 2 away from it, at 1 and at 5. Indeed x^2 - 6x + 5 = (x - 1)(x - 5).
So \displaystyle \sqrt{D} measures how spread out the roots are. Now keep a and b and change only c. The middle \displaystyle -\frac{b}{2a} has no c in it, so it stays at 3. But D = 36 - 4c shrinks as c grows, and the roots close in:
At c = 5 they are 4 apart, at c = 8 they are 2 apart, at c = 9 they meet at 3, and at c = 10 the discriminant is -4. No real number has a square root of -4, so there is no real distance to step out by, and no real roots at all.
The parabola and the axis
The same story on the graph. Take one parabola and slide it straight up: in the next figure the curve is y = x^2 - 2x + c, and only c changes.
With c = -3 the lowest point is below the axis, and since both arms rise for ever, the curve has to come up through the axis once on each side: two crossings. Raise it to c = 1 and the lowest point sits exactly on the axis: one touch. Raise it to c = 3 and the lowest point is above the axis, so nothing on the curve is lower still: no meeting at all. Meanwhile D = b^2 - 4ac = 4 - 4c goes 16, 0, -8. Lifting the parabola pushes D down, and the moment the vertex reaches the axis is the moment D reaches zero.
The same thing happens upside down. A parabola that opens downward has a highest point instead of a lowest one, and it crosses the axis exactly when that highest point is above the axis.
In both directions the rule is the same: the graph crosses when the vertex is on the far side of the axis from the way the parabola opens. The discriminant is a way of measuring that.
A negative discriminant
A negative discriminant is not the end of the problem. Once you have the imaginary unit i, with i^2 = -1, a negative number does have square roots: for d > 0, \displaystyle \sqrt{-d} = i\sqrt{d}, since \displaystyle (i\sqrt{d})^2 = i^2 d = -d. The formula then carries on without a single change. The middle \displaystyle -\frac{b}{2a} is still real; only the step out from it has become imaginary.
x^2 - 2x - 3 = 0 and x^2 - 2x + 5 = 0 differ only in the constant term, so their parabolas share the line of symmetry x = 1. The first crosses the axis 2 either side of it, at 1 \pm 2. The second sits wholly above the axis, and its solutions are 1 \pm 2i: the same center, with the 2 now imaginary. Complex solutions are not points on the graph. They are what the formula still finds when the graph has nothing to show. The two solutions differ only in the sign of their imaginary part, so in the complex plane they are mirror images across the real axis.
Results
What the discriminant is
The formula in terms of D. Every quadratic equation ax^2 + bx + c = 0 with a \ne 0 has roots
Where the roots sit. When D \ge 0, the two roots are symmetric about \displaystyle -\frac{b}{2a}, each at a distance \displaystyle \frac{\sqrt{D}}{2|a|} from it. So they are \displaystyle \frac{\sqrt{D}}{|a|} apart.
What it tells apart. If the two roots are r and s, real or not, then
So D = 0 exactly when r = s: the discriminant is zero precisely when the two roots cannot be told apart, which is the reason for its name.
Rational roots. When a, b and c are integers, the roots are rational exactly when D is a perfect square (0, 1, 4, 9, 16, ...). In that case the quadratic factors into two linear factors with integer coefficients.
The discriminant and the number of roots
For ax^2 + bx + c = 0 with real a, b, c and a \ne 0:
- If D > 0, the equation has two different real roots, \displaystyle x = \frac{-b \pm \sqrt{D}}{2a}.
- If D = 0, it has exactly one real root, \displaystyle x = -\frac{b}{2a}, and \displaystyle ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2.
- If D < 0, it has no real roots. It has two complex roots instead.
If a and c have opposite signs, there are two real roots, whatever b is: ac is negative, so -4ac is positive, and b^2 - 4ac is a number that is at least 0 plus a positive number. The shortcut only runs one way: if a and c have the same sign, the count depends on whether b^2 is big enough to beat 4ac.
The discriminant and the graph
For the graph of y = ax^2 + bx + c:
- If D > 0, it crosses the x-axis at two points.
- If D = 0, it touches the x-axis at one point, and that point is the vertex.
- If D < 0, it does not meet the x-axis.
The height of the vertex. The vertex is at (h, k) with
so D = -4ak. The discriminant is positive exactly when a and k have opposite signs.
The gap between the intercepts. When D > 0, the two x-intercepts are \displaystyle \frac{\sqrt{D}}{|a|} apart, centered on the line of symmetry x = h.
When the formula gives complex roots
If D < 0, write \displaystyle \sqrt{D} = i\sqrt{-D}. The two roots are
a pair p + qi and p - qi with the same real part \displaystyle p = -\frac{b}{2a} and opposite imaginary parts. Such a pair is called a pair of complex conjugates. As long as a, b and c are real, the non-real roots of a quadratic always come as a conjugate pair.
Sum and product. In every case, real or complex, the two roots r and s satisfy
which gives a quick check on any pair of solutions, and a way to build an equation from them.
Why it is true
The reshaping. Multiply ax^2 + bx + c = 0 by 4a, which is allowed because a \ne 0, then take 4ac from both sides and add b^2 to both sides:
The left side is (2ax + b)^2, so the equation is exactly (2ax + b)^2 = D, with the same roots as the original. This is the derivation of the quadratic formula, which is completing the square in disguise. Read as an identity it says 4a(ax^2 + bx + c) = (2ax + b)^2 - D for every x. Taking square roots, \displaystyle 2ax + b = \pm\sqrt{D}, so \displaystyle x = \frac{-b \pm \sqrt{D}}{2a}.
Where the roots sit, and what D tells apart. Call the roots \displaystyle r = \frac{-b + \sqrt{D}}{2a} and \displaystyle s = \frac{-b - \sqrt{D}}{2a}. Their average is \displaystyle -\frac{b}{2a}, the middle. Their difference is \displaystyle r - s = \frac{2\sqrt{D}}{2a} = \frac{\sqrt{D}}{a}, and squaring gives \displaystyle (r - s)^2 = \frac{D}{a^2}, that is, D = a^2(r - s)^2. When D \ge 0 the distance between them is \displaystyle |r - s| = \frac{\sqrt{D}}{|a|}, half of it on each side of the middle.
Rational roots. If D = m^2 for a whole number m, the roots \displaystyle \frac{-b \pm m}{2a} are fractions of integers. If D is a positive whole number that is not a perfect square, \displaystyle \sqrt{D} is irrational, and so is r, since otherwise \displaystyle \sqrt{D} = 2ar + b would be rational.
The count. Put u = 2ax + b. Since a \ne 0, \displaystyle x = \frac{u - b}{2a}, so real values of x and real values of u match one for one, and the count of real roots is the count of real u with u^2 = D. Everything now rests on one fact about squares: the square of a real number is never negative, it is 0 only for 0, and a positive number has exactly two square roots, one positive and one negative.
- If D > 0, u can be \displaystyle \sqrt{D} or \displaystyle -\sqrt{D}, two different numbers, so there are two different roots.
- If D = 0, the only number whose square is 0 is 0. So 2ax + b = 0, and \displaystyle x = -\frac{b}{2a} is the one root. The identity above, with D = 0 and divided by 4a, gives \displaystyle ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2.
- If D < 0, no real u has u^2 = D, so no real x works.
The graph. Complete the square on the right-hand side of y = ax^2 + bx + c:
That is vertex form, y = a(x - h)^2 + k, with \displaystyle h = -\frac{b}{2a} and \displaystyle k = -\frac{D}{4a}. Suppose a > 0. The squared term is never negative, so the lowest value of y is k, reached at x = h, and the parabola rises without limit on both sides. If D > 0 then k < 0: the lowest point is below the axis and the arms climb above it, so the curve crosses once on each side of x = h. If D = 0 then k = 0: the lowest point is on the axis and every other point is above it. If D < 0 then k > 0: even the lowest point is above the axis. If a < 0, k is the highest value and \displaystyle k = \frac{D}{-4a} has the same sign as D, so a positive D puts the highest point above the axis with the arms falling below it, and the three cases come out the same way.
Complex roots. If D < 0, then \displaystyle (i\sqrt{-D})^2 = i^2(-D) = D, so \displaystyle u = \pm i\sqrt{-D} both square to D, and \displaystyle x = \frac{-b \pm i\sqrt{-D}}{2a}. Both have real part \displaystyle -\frac{b}{2a}, and their imaginary parts differ only in sign, which makes them conjugates.
Sum and product. If r and s are the two roots, the quadratic factors as a(x - r)(x - s) = ax^2 - a(r + s)x + ars. Matching coefficients with ax^2 + bx + c gives -a(r + s) = b and ars = c.
Worked examples
Counting without solving
How many real roots does each equation have? Once it is in standard form ax^2 + bx + c = 0, how does the parabola y = ax^2 + bx + c meet the x-axis?
{\text{(a)}\;\; 2x^2 - 7x + 4 = 0} \qquad {\text{(b)}\;\; 9x^2 - 12x + 4 = 0} \qquad {\text{(c)}\;\; 3x^2 + 2 = -4x}
SolutionRead off a, b and c, signs included, and work out b^2 - 4ac. Only the sign of the answer matters.
(a) a = 2, b = -7, c = 4:
The bracket round -7 matters: (-7)^2 = 49, whereas -7^2 would mean -49. Since 17 > 0, there are two real roots, and the parabola y = 2x^2 - 7x + 4 crosses the x-axis twice.
(b) a = 9, b = -12, c = 4:
The discriminant is 0, so there is one real root, \displaystyle x = -\frac{-12}{2 \cdot 9} = \frac{2}{3}, and the parabola touches the axis at its vertex, \displaystyle \left(\frac{2}{3}, 0\right). The left side is the perfect square (3x - 2)^2, which is why.
(c) This one is not in standard form, and its a, b, c mean nothing until it is. Add 4x to both sides: 3x^2 + 4x + 2 = 0. Now a = 3, b = 4, c = 2:
-8 < 0, so there are no real roots. Since a = 3 > 0 the parabola y = 3x^2 + 4x + 2 opens upward, so it lies entirely above the axis.
Squaring b always gives a number that is 0 or more, so a negative b never makes the first term negative. The sign of the whole discriminant comes from how b^2 compares with 4ac.
Standard form first, from either side
Find the discriminant of 5 - 2x^2 = 3x.
SolutionThe terms are split across both sides, so gather them first. Moving everything to the right gives
so a = 2, b = 3, c = -5 and
Moving everything to the left instead gives -2x^2 - 3x + 5 = 0, with a = -2, b = -3, c = 5:
The same number, and not by luck. Moving everything to the other side multiplies every term by -1. That leaves b^2 alone, and it flips the signs of both a and c, which leaves their product ac alone. So it does not matter which side you gather on, only that you gather.
Since 49 = 7^2 is a perfect square, the roots are rational: 2x^2 + 3x - 5 = (2x + 5)(x - 1).
A missing term, and fractions
Find the discriminant of each equation.
{\text{(a)}\;\; 3x^2 - 12 = 0} \qquad {\text{(b)}\;\; 2x^2 + 5x = 0} \qquad \displaystyle {\text{(c)}\;\; \frac{1}{2}x^2 - x + \frac{1}{3} = 0}
Solution(a) There is no x term, so b = 0, with a = 3 and c = -12:
The roots are \pm 2, and the distance between them is \displaystyle \frac{\sqrt{144}}{3} = 4, as the Results say it must be.
(b) There is no constant term, so c = 0:
(c) The coefficients are fractions, which changes nothing about the definition. \displaystyle a = \frac{1}{2}, b = -1, \displaystyle c = \frac{1}{3}:
If you clear the fractions first by multiplying through by 6, the equation becomes 3x^2 - 6x + 2 = 0 with D = 36 - 24 = 12. That is 36 = 6^2 times as big: multiplying every term by 6 multiplies b^2 by 36 and 4ac by 36 as well. The roots are the same, and so is the sign of D.
Opposite signs settle it
Show that 3x^2 + 17x - 250 = 0 has two real roots, using only the signs of a and c.
SolutionHere a = 3 is positive and c = -250 is negative, so ac = -750 is negative. Then -4ac = 3000 is positive. The discriminant is
and b^2, being a square, is at least 0. So D is at least 3000, which is positive, and there are two real roots. The value of b never mattered. (For the record, D = 289 + 3000 = 3289.)
On the graph the same fact is plain: the y-intercept (0, -250) is below the axis while the parabola opens upward, so it must rise through the axis on each side.
Will it factor?
Decide which of these factors with integer coefficients, and factor the one that does.
{\text{(a)}\;\; 6x^2 + x - 15} \qquad {\text{(b)}\;\; 6x^2 + x - 14}
SolutionTrying every factor pair of 6 against every factor pair of the constant is slow. The discriminant answers the question in one line.
(a) Here D = 1^2 - 4(6)(-15), which is 1 + 360 = 361 = 19^2, a perfect square, so it factors. The roots are
that is, \displaystyle x = \frac{18}{12} = \frac{3}{2} or \displaystyle x = \frac{-20}{12} = -\frac{5}{3}.
The root \displaystyle \frac{3}{2} comes from a factor 2x - 3, and the root \displaystyle -\frac{5}{3} from a factor 3x + 5. Multiplying out, (2x - 3)(3x + 5) = 6x^2 + 10x - 9x - 15, which is 6x^2 + x - 15.
(b) D = 1^2 - 4(6)(-14) = 1 + 336 = 337. Since 18^2 = 324 and 19^2 = 361, 337 is not a perfect square. The roots are irrational, so no factoring with integer coefficients exists, and no amount of trying factor pairs will find one.
The vertex and the intercepts from one number
For y = x^2 - 6x + 6, find the discriminant, the vertex and the x-intercepts, and check them against each other.
SolutionD = (-6)^2 - 4(1)(6) = 36 - 24 = 12. It is positive, so expect two x-intercepts.
The vertex is at \displaystyle h = -\frac{-6}{2(1)} = 3 and \displaystyle k = -\frac{D}{4a} = -\frac{12}{4} = -3. As a check, substitute: 3^2 - 6(3) + 6 = 9 - 18 + 6 = -3. The vertex (3, -3) is below the axis and the parabola opens upward, which agrees with two crossings.
The intercepts sit either side of the middle h = 3, each \displaystyle \frac{\sqrt{D}}{2|a|} = \frac{\sqrt{12}}{2} = \sqrt{3} away, so they are at
about (1.27, 0) and (4.73, 0). They are \displaystyle \frac{\sqrt{D}}{|a|} = \sqrt{12} = 2\sqrt{3} apart.
Check. The formula gives \displaystyle \frac{6 \pm \sqrt{12}}{2} = \frac{6 \pm 2\sqrt{3}}{2}, which is the same pair.
A parabola that opens downward
For y = -2x^2 + 4x + 1, decide how many x-intercepts there are, then find them.
Solutiona = -2, b = 4, c = 1, so
Positive, so two intercepts. It is worth seeing why from the picture: the vertex is at \displaystyle h = -\frac{4}{2(-2)} = 1, with \displaystyle k = -\frac{24}{4(-2)} = 3. The parabola opens downward from a highest point 3 units above the axis, so it must come down through the axis on both sides.
The formula gives
so the intercepts are at \displaystyle x = 1 \pm \frac{\sqrt{6}}{2}, about 2.22 and -0.22. The gap between them is \displaystyle \sqrt{6}, and \displaystyle \frac{\sqrt{D}}{|a|} = \frac{\sqrt{24}}{2} = \frac{2\sqrt{6}}{2} = \sqrt{6}.
From the graph back to the sign
A parabola opens downward and its vertex is at (-1, -2). Is its discriminant positive, zero or negative?
SolutionUse D = -4ak. Opening downward means a < 0, and the vertex gives k = -2. So
which is negative because a is. The discriminant is negative, and the graph has no x-intercepts. That is also plain from the picture: the highest point is already below the axis.
For one particular such parabola, take a = -3: y = -3(x + 1)^2 - 2 = -3x^2 - 6x - 5. Its discriminant is (-6)^2 - 4(-3)(-5) = 36 - 60 = -24, which is 8a = 8(-3), as predicted.
Choosing the constant
For which values of c does the graph of y = x^2 + 6x + c (a) touch the x-axis, (b) cross it twice, (c) miss it?
SolutionThe discriminant is D = 6^2 - 4(1)(c) = 36 - 4c, and each case is a condition on it.
(a) Touching means D = 0: 36 - 4c = 0, so c = 9. Then the curve is y = x^2 + 6x + 9 = (x + 3)^2, which touches the axis at (-3, 0).
(b) Two crossings means D > 0: 36 - 4c > 0, so c < 9.
(c) No meeting means D < 0: 36 - 4c < 0, so c > 9.
This is the sliding parabola of the Intuition again. Changing c moves the graph straight up and down, and c = 9 is the one height at which its lowest point rests on the axis. The vertex height confirms it: \displaystyle k = -\frac{36 - 4c}{4} = c - 9, which is zero exactly when c = 9.
A parameter in two places
For which values of k does x^2 + kx + k + 3 = 0 have two real roots, one, or none?
SolutionHere a = 1, b = k and c = k + 3, so
The discriminant is itself a quadratic, in k, and its sign decides everything. Factoring makes the sign easy to read. A product is 0 exactly when a factor is, so D = 0 when k = 6 or k = -2: one real root. A product is positive when both factors have the same sign, which happens when k > 6 (both positive) or k < -2 (both negative): two real roots. Between, for -2 < k < 6, the factors have opposite signs and D < 0: no real roots, and the two roots are a complex conjugate pair.
Check. At k = 6 the equation is x^2 + 6x + 9 = (x + 3)^2 = 0, root -3. At k = -2 it is x^2 - 2x + 1 = (x - 1)^2 = 0, root 1. At k = 0, inside the gap, it is x^2 + 3 = 0, which plainly has no real root.
When the leading coefficient is the unknown
For which values of k does kx^2 - 6x + 3 = 0 have no real roots? For which does it have two?
SolutionHere a = k, b = -6, c = 3:
No real roots means 36 - 12k < 0, so 12k > 36 and k > 3.
Two real roots means 36 - 12k > 0, so k < 3. But one value below 3 must be thrown out: k = 0. The theorem needs a \ne 0, and when k = 0 the equation is -6x + 3 = 0, which is not quadratic at all. It has the single root \displaystyle x = \frac{1}{2}, although the discriminant formula would give 36 and claim two. So the answer is k < 3 with k \ne 0.
At the boundary, k = 3 gives 3x^2 - 6x + 3 = 3(x - 1)^2 = 0, with one root, x = 1.
Choosing a coefficient to control the roots
(a) For which values of k does x^2 + kx + 9 = 0 have its two roots equal?
(b) For which value of c are the roots of 2x^2 - 5x + c = 0 exactly \displaystyle \frac{3}{2} apart?
Solution(a) Since D = a^2(r - s)^2, the roots are equal exactly when D = 0:
and k = 6 or k = -6. With k = 6 the equation is (x + 3)^2 = 0, with the single root -3. With k = -6 it is (x - 3)^2 = 0, with the single root 3.
(b) The roots are \displaystyle \frac{\sqrt{D}}{|a|} apart, and here D = 25 - 8c and |a| = 2:
and squaring, 25 - 8c = 9, so c = 2. Squaring was safe because both sides were already non-negative.
Check. 2x^2 - 5x + 2 = (2x - 1)(x - 2) has roots \displaystyle \frac{1}{2} and 2, and \displaystyle 2 - \frac{1}{2} = \frac{3}{2}.
Solving when the discriminant is negative
Solve x^2 - 2x + 5 = 0 over the complex numbers.
SolutionHere a = 1, b = -2, c = 5. Work out the discriminant on its own first, since it decides what kind of answer is coming:
It is negative, so there are no real solutions, and the graph of y = x^2 - 2x + 5 never meets the x-axis. If the question is about real solutions, that is a complete answer. Over the complex numbers, rewrite the square root of the negative number with i, as square roots of negative numbers are defined: \displaystyle \sqrt{-16} = i\sqrt{16} = 4i. Substitute:
A complex number is written p + qi, with its real part and its imaginary part kept apart. Dividing a sum by 2 divides each of its terms by 2, so
Check. Substituting works, using i^2 = -1 wherever it appears. Since (1 + 2i)^2 = 1 + 4i + 4i^2 = -3 + 4i,
The sum and product check needs no squaring: the sum is \displaystyle (1 + 2i) + (1 - 2i) = 2 = -\frac{-2}{1}, and the product is \displaystyle (1 + 2i)(1 - 2i) = 1 - 4i^2 = 5 = \frac{5}{1}. Both match.
Fractions, radicals and a missing term
Solve each equation over the complex numbers.
{\text{(a)}\;\; 2x^2 - 2x + 5 = 0} \qquad {\text{(b)}\;\; 5x^2 + 4x + 2 = 0} \qquad {\text{(c)}\;\; 4x^2 + 9 = 0}
Solution(a) Here D = (-2)^2 - 4(2)(5) = 4 - 40 = -36, so \displaystyle \sqrt{-36} = 6i, and the denominator is 2a = 4:
Splitting the fraction first lets each part reduce on its own. Nothing requires the real part and the imaginary part to end up with the same denominator. Check: the sum is \displaystyle 1 = -\frac{-2}{2}, and the product is \displaystyle \frac{1}{4} + \frac{9}{4} = \frac{5}{2}, which is \displaystyle \frac{c}{a}.
(b) Here D = 4^2 - 4(5)(2) = 16 - 40 = -24. Take the i out first, then simplify the radical as usual. Since 24 = 4 \cdot 6, \displaystyle \sqrt{-24} = i\sqrt{24} = 2i\sqrt{6}, and dividing each term by 10,
Writing \displaystyle i\sqrt{6}, with the i in front, keeps the i from being read as part of the radicand. Check: the sum is \displaystyle -\frac{4}{5}, and the product is \displaystyle \frac{4}{25} + \frac{6}{25} = \frac{2}{5}.
(c) There is no x term, so b = 0, and D = 0 - 4(4)(9) = -144, with \displaystyle \sqrt{-144} = 12i:
The real part is 0, because the line of symmetry is \displaystyle x = -\frac{b}{2a} = 0, the y-axis. The square root property gets there faster: \displaystyle x^2 = -\frac{9}{4}, so \displaystyle x = \pm\frac{3}{2}i. Two methods, one answer, as there must be.
A product to expand first
Solve x(x + 4) = -7 over the complex numbers.
SolutionThis is a product equal to -7, not a quadratic equal to 0, so a, b and c cannot be read off yet. Expand, then move the -7 across:
Now a = 1, b = 4 and c = 7, so D = 16 - 28 = -12 and \displaystyle \sqrt{-12} = i\sqrt{12} = 2i\sqrt{3}:
Check in the original equation. With \displaystyle x = -2 + i\sqrt{3}, the second factor is \displaystyle x + 4 = 2 + i\sqrt{3}, and the product of the two is a difference of squares:
Building a quadratic from its solutions
Find a quadratic equation x^2 + bx + c = 0, with integer coefficients, whose solutions are 3 \pm 2i.
SolutionRun the sum and product check backwards. With a = 1, the sum of the solutions is -b and their product is c. The sum is (3 + 2i) + (3 - 2i) = 6, so b = -6. The product is
so c = 13, and the equation is x^2 - 6x + 13 = 0.
Check with the formula: D = 36 - 52 = -16, so \displaystyle x = \frac{6 \pm 4i}{2} = 3 \pm 2i. The same equation comes from multiplying out (x - 3 - 2i)(x - 3 + 2i), as in writing a polynomial from its zeros.
Where the solutions turn complex
For which values of k does x^2 + kx + 9 = 0 have non-real solutions?
SolutionThe solutions are non-real exactly when the discriminant is negative. Here a = 1, b = k and c = 9, so
At the two ends the discriminant is 0: these are the equal roots of Example 12(a), -3 when k = 6 and 3 when k = -6. For k outside the interval there are two real solutions. Inside it the solutions are a conjugate pair; at k = 0, for instance, they are \pm 3i.
On the graph, the vertex of y = x^2 + kx + 9 is at \displaystyle \left(-\frac{k}{2}, 9 - \frac{k^2}{4}\right). As k moves away from 0 the vertex slides sideways and down, and k = \pm 6 is the moment it lands on the axis.
When does a line touch a parabola?
For which value of m does the line y = 2x + m touch the parabola y = x^2? Where do they touch?
SolutionA point on both graphs has the same y on both, so set the two right-hand sides equal:
Each real solution of this equation is the x-coordinate of a meeting point, so the question about two graphs has become a question about one quadratic. Its discriminant is
The line touches the parabola when there is exactly one meeting point, D = 0, so m = -1. Then x^2 - 2x + 1 = (x - 1)^2 = 0 gives x = 1, and the point is (1, 1).
For m > -1 the line cuts the parabola twice, and for m < -1 it misses. A line that meets a parabola at exactly one point without crossing into it is its tangent there, so this found the tangent line y = 2x - 1 to y = x^2 at (1, 1), with no calculus.
Two numbers from their sum and product
Show that two real numbers with sum s and product p exist exactly when s^2 \ge 4p. Then decide whether two real numbers can have sum 10 and product 21, or sum 10 and product 26.
SolutionSuppose r and t are such numbers. Then t = s - r, and rt = p becomes r(s - r) = p, that is
Conversely, if r is any real root of this equation, put t = s - r. Then r + t = s and rt = r(s - r) = p. So the numbers exist exactly when this quadratic in r has a real root, that is when
When D = 0 the one root is \displaystyle r = \frac{s}{2}, and then \displaystyle t = \frac{s}{2} as well: the two numbers are equal.
Sum 10, product 21: D = 100 - 84 = 16 > 0, so yes. The equation r^2 - 10r + 21 = 0 factors as (r - 3)(r - 7) = 0, and the numbers are 3 and 7.
Sum 10, product 26: D = 100 - 104 = -4 < 0, so no real pair exists.
Read the other way, s^2 \ge 4p says \displaystyle p \le \frac{s^2}{4}: two numbers with sum s have product at most \displaystyle \frac{s^2}{4}, reached only when both are \displaystyle \frac{s}{2}. With sum 10, the largest product is 25, from 5 and 5. A fact about all pairs of numbers has come out of counting the roots of one quadratic.
Showing an expression is always positive
Show that x^2 - 3x + 4 > 0 for every real number x, and find its smallest value.
SolutionThink of it as the graph y = x^2 - 3x + 4. Its discriminant is (-3)^2 - 4(1)(4) = 9 - 16 = -7, which is negative, so the graph never meets the x-axis. Since a = 1 > 0 it opens upward, and a graph that opens upward and never reaches the axis lies entirely above it. So y > 0 for every x.
The smallest value is the height of the vertex:
at \displaystyle x = -\frac{-3}{2} = \frac{3}{2}. Check: \displaystyle \left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) + 4 = \frac{9}{4} - \frac{18}{4} + \frac{16}{4} = \frac{7}{4}.
Boundaries
The equation must be quadratic. If a = 0 there is no x^2 term and b^2 - 4ac says nothing. 3x + 1 = 0 has one root, yet b^2 - 4ac = 9 would claim two; the graph y = 3 has D = 0 and no x-intercept at all. Whenever a contains an unknown, as in Example 11, the value that makes a = 0 has to be looked at separately.
The equation must be in standard form. The letters a, b, c are read from ax^2 + bx + c = 0 only, with their signs. Reading 3x^2 = 2x + 5 as it stands, with b = 2 and c = 5, gives 4 - 60 = -56 and would suggest no real roots. The true discriminant, from 3x^2 - 2x - 5 = 0, is 64.
Its size belongs to the equation, not only to the roots. Multiplying the equation by k multiplies D by k^2 and leaves the roots alone. So the sign of D means the same in every version, but a size such as “D = 12” only means something next to its a. What stays fixed is \displaystyle \frac{\sqrt{D}}{2|a|}.
The perfect square test needs integer coefficients. \displaystyle x^2 - 2\sqrt{2}\,x + 1 = 0 has D = 8 - 4 = 4, a perfect square, yet its roots are \displaystyle \sqrt{2} \pm 1, which are irrational. With rational coefficients, clear the fractions first.
The sign, and the conjugate pair, need real coefficients. The argument used the fact that a real square is never negative. With complex coefficients the discriminant can itself be complex and have no sign at all, and even D = 0 need not give a real root: x^2 + 2ix - 1 = 0 has D = 0 and its one root is -i. The non-real roots need not be conjugates either: x^2 - (1 + i)x + i = 0 factors as (x - 1)(x - i) = 0, with roots 1 and i. The formula and D = a^2(r - s)^2 still hold; only the reading by sign stops.
One root means one number, not one factor. When D = 0 the root comes from a squared factor, so in counts that include multiplicity it is counted twice. That is how every quadratic ends up with exactly two roots, counted properly, over the complex numbers.
The real question and the complex question have different answers. Over the real numbers, x^2 - 2x + 5 = 0 has no solution, and saying so is correct. The complex solutions 1 \pm 2i are not x-intercepts: a graph drawn on real axes can only show real roots.
Take the i out before using any other radical rule. The rule \displaystyle \sqrt{a}\sqrt{b} = \sqrt{ab} holds when a and b are not negative, and fails when both are negative: \displaystyle \sqrt{-4}\sqrt{-9} = 2i \cdot 3i = 6i^2 = -6, while \displaystyle \sqrt{(-4)(-9)} = \sqrt{36} = 6. Writing \displaystyle \sqrt{-d} as \displaystyle i\sqrt{d} first keeps every later step on positive numbers.
The sign counts; it does not locate. The sign of D says how many roots, not where. Two roots could both be positive, both negative, or one of each, and two parabolas with the same D can sit in quite different places. The discriminant is also about the x-axis only: every parabola y = ax^2 + bx + c meets the y-axis exactly once, at (0, c), whatever D is. To ask about any other line, first rewrite the question as one quadratic set equal to zero, as in Example 18.
Where this goes next
Once complex solutions are allowed, every quadratic equation has solutions: two different ones, or one repeated. That is the degree-two case of the fundamental theorem of algebra. The pairing of p + qi with p - qi is not special to quadratics either: any polynomial with real coefficients has its non-real roots in conjugate pairs, which is why a cubic with real coefficients always has at least one real root.
The perfect square test feeds straight into choosing how to solve a quadratic: a perfect square D says factoring will work, anything else says reach for the formula. And the vertex height \displaystyle k = -\frac{D}{4a} is the bridge to graphing a quadratic from standard form.
Example 18 and Example 19 turned questions about two curves, or two numbers, into a question about the roots of one quadratic. The same habit runs through nonlinear systems and the intersection of a line and a circle, where a discriminant of 0 is exactly the condition for a tangent line. Example 10 and Example 20, where the sign of a quadratic had to be read over a whole range, are the start of quadratic inequalities. The expression B^2 - 4AC built from a general second-degree equation in x and y tells an ellipse from a parabola from a hyperbola.
And a negative discriminant has a meaning well outside algebra. The eigenvalues of a 2 \times 2 matrix are the solutions of a quadratic, and when they come out as a complex pair, the matrix turns every direction in the plane. The same complex pair, in a spring or a pendulum, is what makes the motion swing back and forth, as in simple harmonic motion.
Exercises
Count the real roots
Find the discriminant, and say how many real roots the equation has.
- x^2 + 5x + 3 = 0
- 9x^2 - 30x + 25 = 0
- 2x^2 + 3x + 5 = 0
- 5x^2 - 7 = 0
- 3x^2 - 4x = 0
- \displaystyle \frac{1}{3}x^2 - 2x + 3 = 0
- 4x^2 = 3x - 1
- 4 - x = 3x^2
- x(x + 6) = -10
- (x + 1)^2 = 3x
How does the graph meet the axis?
Find the discriminant, and say whether the graph crosses the x-axis twice, touches it once or misses it.
- y = x^2 + 3x - 10
- y = -x^2 + 4x - 5
- y = -3x^2 + 2x + 1
- y = 4x^2 + 12x + 9
Where the roots sit
- For x^2 - 10x + 16 = 0, find D, the middle \displaystyle -\frac{b}{2a}, and the two roots.
- For 3x^2 + 6x - 1 = 0, find D, the middle \displaystyle -\frac{b}{2a}, and the two roots.
For each graph, find the discriminant, the vertex and every x-intercept.
- y = x^2 - 6x + 4
- y = -x^2 + 2x + 2
- y = 2x^2 - 8x + 8
- y = x^2 + 2x + 5
- Without solving, find how far apart the roots of 2x^2 - 3x - 7 = 0 are.
- A quadratic equation with a = 3 has roots 1 and 5. Find its discriminant without writing the equation down.
- Does 4x^2 + 4x - 15 factor with integer coefficients? Does 4x^2 + 4x - 13? Factor any that do.
Complex roots
- Write \displaystyle \sqrt{-18} and \displaystyle \sqrt{-75} in terms of i, with the radicals simplified.
Solve over the complex numbers. Give non-real solutions in the form p \pm qi.
- x^2 + 4x + 5 = 0
- -x^2 + 2x - 10 = 0
- 3x^2 - 2x + 1 = 0
- 9x^2 + 16 = 0
- x(x - 2) = -4
- (x + 3)^2 = 2x - 1
- \displaystyle \frac{1}{2}x^2 + x + 3 = 0
- Find an equation x^2 + bx + c = 0, with integer coefficients, whose solutions are 2 \pm 5i.
- For which values of c does x^2 + 4x + c = 0 have non-real solutions?
Find the coefficient
- For which values of k does x^2 + kx + 25 = 0 have exactly one real root?
- For what value of c does the graph of y = x^2 - 8x + c touch the x-axis, and where?
- For which values of c does 2x^2 + 5x + c = 0 have no real roots?
- For which values of c does the graph of y = x^2 + 5x + c cross the x-axis twice?
- For which values of b does the graph of y = x^2 + bx + 9 miss the x-axis?
- For which values of a does the graph of y = ax^2 + 4x + 1 have no x-intercepts?
- For which values of k does kx^2 + 6x + 1 = 0 have exactly one real solution?
- Find c so that the roots of x^2 - 6x + c = 0 are 4 apart.
- Show that x^2 + (k + 1)x + k = 0 has at least one real root for every k, and find the value of k for which it has exactly one.
- For what value of c does the line y = 4x + c touch the parabola y = x^2 + 1, and at which point?
- How many times does the line y = 5 meet the parabola y = x^2 - 2x + 3? How many times does the line y = 2 meet it?
From the graph
- A parabola opens upward and its vertex is at (3, -4). Is its discriminant positive, zero or negative?
- A parabola opens downward and its vertex is at (-2, -1). Is its discriminant positive, zero or negative?
- The parabola y = a(x - 1)^2 + 2 has discriminant 40. Find a.
Using the count
- A ball’s height after t seconds is h = -16t^2 + 64t + 5 feet. Use the discriminant to decide how many times the ball is at a height of (a) 60 feet, (b) 69 feet, (c) 80 feet.
- For which values of p are there two real numbers with sum 8 and product p? For which p are the two numbers equal, and what are they?
- Is there a rectangle with perimeter 20 and area 30? Use the discriminant to decide.
Reasoning
- Show that if b^2 = 4ac, then \displaystyle ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2.
- Show that no quadratic equation with whole-number coefficients has discriminant 2.
- Explain why ax^2 + bx + c = 0 and cx^2 + bx + a = 0, with a and c both non-zero, always have the same number of real roots.
- Show that (x - p)(x - q) = 1 has two real roots for every pair of real numbers p and q.
- Explain why multiplying every term of a quadratic equation by -1 does not change its discriminant, and say what multiplying every term by 3 does to it.
- Can a quadratic equation with integer coefficients have one rational root and one irrational root? Use the discriminant to explain, and say why D = 0 always gives a rational root.
- Suppose a > 0. Adding 1 to c lifts the graph of y = ax^2 + bx + c by one unit. Show that it makes the discriminant smaller, and say what that does to the x-intercepts.
- Explain, using the vertex, why a parabola whose discriminant is 0 meets the x-axis at exactly one point.
- One solution of a quadratic equation with real coefficients is 3 - 4i. What is the other, and why? What is the x-coordinate of the vertex of its graph?
- Work out \displaystyle \sqrt{-4} \cdot \sqrt{-9} by writing each root in terms of i first. Is it equal to \displaystyle \sqrt{(-4)(-9)}? Explain.
Answers
- Two. D = 5^2 - 4(1)(3) = 25 - 12 = 13 > 0.
- One. D = (-30)^2 - 4(9)(25) = 900 - 900 = 0. The root is \displaystyle x = \frac{5}{3}, since the left side is (3x - 5)^2.
- None. D = 3^2 - 4(2)(5) = 9 - 40 = -31 < 0.
- Two. With b = 0: D = 0^2 - 4(5)(-7) = 140 > 0.
- Two. With c = 0: D = (-4)^2 - 4(3)(0) = 16 > 0. The roots are x = 0 and \displaystyle x = \frac{4}{3}, since 3x^2 - 4x = x(3x - 4).
- One. \displaystyle D = (-2)^2 - 4\left(\frac{1}{3}\right)(3) = 4 - 4 = 0. The root is x = 3.
- None. In standard form 4x^2 - 3x + 1 = 0, and D = 9 - 16 = -7 < 0.
- Two. In standard form 3x^2 + x - 4 = 0, and D = 1^2 - 4(3)(-4) = 1 + 48 = 49 > 0. Since 49 = 7^2 the roots are rational: x = 1 and \displaystyle x = -\frac{4}{3}.
- None. Expand and move the -10 across: x^2 + 6x + 10 = 0, and D = 36 - 40 = -4 < 0.
- None. Expand and move the 3x across: x^2 + 2x + 1 - 3x = x^2 - x + 1 = 0, and D = 1 - 4 = -3 < 0.
- D = 49, crosses twice. 3^2 - 4(1)(-10) = 9 + 40 = 49; the intercepts are x = 2 and x = -5.
- D = -4, misses. 16 - 4(-1)(-5) = 16 - 20; it opens downward, so it lies below the axis.
- D = 16, crosses twice. 4 - 4(-3)(1) = 4 + 12; the intercepts are x = 1 and \displaystyle x = -\frac{1}{3}.
- D = 0, touches once, at \displaystyle \left(-\frac{3}{2}, 0\right). 144 - 144 = 0, and the expression is (2x + 3)^2.
- D = 36, the middle is 5, and the roots are x = 2 and x = 8. Each root is \displaystyle \frac{\sqrt{36}}{2} = 3 from the middle.
- D = 48, the middle is -1, and the roots are \displaystyle x = \frac{-3 \pm 2\sqrt{3}}{3}. Here D = 36 + 12 = 48, and each root is \displaystyle \frac{\sqrt{48}}{6} = \frac{4\sqrt{3}}{6} = \frac{2\sqrt{3}}{3} from -1.
- D = 20, vertex (3, -5), intercepts \displaystyle x = 3 \pm \sqrt{5}. \displaystyle k = -\frac{20}{4} = -5, and each intercept is \displaystyle \frac{\sqrt{20}}{2} = \sqrt{5} from x = 3.
- D = 12, vertex (1, 3), intercepts \displaystyle x = 1 \pm \sqrt{3}. D = 4 + 8, \displaystyle k = -\frac{12}{4(-1)} = 3, and \displaystyle x = \frac{-2 \pm \sqrt{12}}{-2} = 1 \mp \sqrt{3}, which is the same pair.
- D = 0, vertex (2, 0), one intercept, x = 2. D = 64 - 64 = 0, and 2x^2 - 8x + 8 = 2(x - 2)^2: the vertex is on the axis.
- D = -16, vertex (-1, 4), no x-intercepts. D = 4 - 20 and \displaystyle k = -\frac{-16}{4} = 4: the lowest point is 4 above the axis.
- \displaystyle \frac{\sqrt{65}}{2} apart. D = 9 + 56 = 65, and the distance is \displaystyle \frac{\sqrt{D}}{|a|}.
- D = 144, since D = a^2(r - s)^2 = 9(1 - 5)^2 = 9 \cdot 16.
- The first does: (2x - 3)(2x + 5). Its D = 16 + 240 = 256 = 16^2, and the roots are \displaystyle \frac{-4 \pm 16}{8}, that is \displaystyle \frac{3}{2} and \displaystyle -\frac{5}{2}. The second does not: its D = 16 + 208 = 224 lies between 14^2 = 196 and 15^2 = 225, so it is not a perfect square.
- \displaystyle \sqrt{-18} = i\sqrt{9 \cdot 2} = 3i\sqrt{2} and \displaystyle \sqrt{-75} = i\sqrt{25 \cdot 3} = 5i\sqrt{3}.
- x = -2 \pm i. D = 16 - 20 = -4, so \displaystyle x = \frac{-4 \pm 2i}{2}.
- x = 1 \pm 3i. Here a = -1, b = 2, c = -10, and D = 4 - 40 = -36, so \displaystyle x = \frac{-2 \pm 6i}{-2} = 1 \mp 3i, which is the same pair. Multiplying the equation by -1 first, to get x^2 - 2x + 10 = 0, gives it too.
- \displaystyle x = \frac{1}{3} \pm \frac{\sqrt{2}}{3}i. D = 4 - 12 = -8 and \displaystyle \sqrt{-8} = 2i\sqrt{2}, so \displaystyle x = \frac{2 \pm 2i\sqrt{2}}{6}.
- \displaystyle x = \pm\frac{4}{3}i. Here b = 0: D = -4(9)(16) = -576 and \displaystyle \sqrt{-576} = 24i, so \displaystyle x = \pm\frac{24i}{18}.
- \displaystyle x = 1 \pm i\sqrt{3}. Expand and move the -4 across: x^2 - 2x + 4 = 0. D = 4 - 16 = -12 and \displaystyle \sqrt{-12} = 2i\sqrt{3}, so \displaystyle x = \frac{2 \pm 2i\sqrt{3}}{2}.
- \displaystyle x = -2 \pm i\sqrt{6}. Expand and collect: x^2 + 4x + 10 = 0. D = 16 - 40 = -24 and \displaystyle \sqrt{-24} = 2i\sqrt{6}, so \displaystyle x = \frac{-4 \pm 2i\sqrt{6}}{2}.
- \displaystyle x = -1 \pm i\sqrt{5}. Multiply by 2 to clear the fraction: x^2 + 2x + 6 = 0. D = 4 - 24 = -20 and \displaystyle \sqrt{-20} = 2i\sqrt{5}, so \displaystyle x = \frac{-2 \pm 2i\sqrt{5}}{2}.
- x^2 - 4x + 29 = 0. The sum of the solutions is 4 = -b and their product is (2 + 5i)(2 - 5i) = 4 + 25 = 29 = c.
- c > 4. The discriminant 16 - 4c is negative exactly when c > 4.
- k = 10 or k = -10. D = k^2 - 100 = 0 gives k^2 = 100.
- c = 16, touching at (4, 0). D = 64 - 4c = 0, and then y = x^2 - 8x + 16 = (x - 4)^2.
- \displaystyle c > \frac{25}{8}. D = 25 - 8c < 0 gives 8c > 25.
- \displaystyle c < \frac{25}{4}. D = 25 - 4c > 0 gives 4c < 25.
- -6 < b < 6. D = b^2 - 36 < 0, so b^2 < 36.
- a > 4. D = 16 - 4a < 0 gives a > 4. Every such a is positive, so the graph opens upward and lies above the axis. (a = 0 is not a parabola: y = 4x + 1 is a line with an x-intercept, so it is rightly left out.)
- k = 9 or k = 0. For k \ne 0 the equation is quadratic, and D = 36 - 4k = 0 gives k = 9, with root \displaystyle x = -\frac{1}{3}. For k = 0 it is the linear equation 6x + 1 = 0, which also has exactly one solution, \displaystyle x = -\frac{1}{6}.
- c = 5. The roots are \displaystyle \sqrt{36 - 4c} apart, so 36 - 4c = 16. The roots are 1 and 5.
- k = 1. D = (k + 1)^2 - 4k = k^2 - 2k + 1 = (k - 1)^2, a square, so it is never negative and there is always a real root. It is 0 only when k = 1, where the equation is (x + 1)^2 = 0.
- c = -3, touching at (2, 5). Setting x^2 + 1 = 4x + c gives x^2 - 4x + (1 - c) = 0, with D = 16 - 4(1 - c) = 12 + 4c = 0. Then (x - 2)^2 = 0, and y = 2^2 + 1 = 5.
- y = 5 meets it twice; y = 2 meets it once, at (1, 2). x^2 - 2x + 3 = 5 gives x^2 - 2x - 2 = 0 with D = 12 > 0. x^2 - 2x + 3 = 2 gives x^2 - 2x + 1 = 0 with D = 0: the line passes through the vertex.
- Positive. D = -4ak = -4a(-4) = 16a, and a > 0. The lowest point is below the axis.
- Negative. D = -4ak = -4a(-1) = 4a, and a < 0. The highest point is below the axis.
- a = -5. D = -4ak = -4a(2) = -8a = 40. Check: y = -5x^2 + 10x - 3 and 100 - 4(-5)(-3) = 100 - 60 = 40.
- (a) Twice, at t = 1.25 and t = 2.75: 16t^2 - 64t + 55 = 0 has D = 4096 - 3520 = 576 > 0. (b) Once, at t = 2: 16t^2 - 64t + 64 = 0 has D = 4096 - 4096 = 0; this is the top of the flight. (c) Never: 16t^2 - 64t + 75 = 0 has D = 4096 - 4800 = -704 < 0.
- p \le 16; the numbers are equal when p = 16, and both are 4. The numbers are the roots of r^2 - 8r + p = 0, with D = 64 - 4p \ge 0.
- No. If one side is x, the other is 10 - x, and x(10 - x) = 30 becomes x^2 - 10x + 30 = 0, with D = 100 - 120 = -20 < 0. No real side length works, as Example 19 predicts: sides with sum 10 have product at most 25.
- Expand the right side: \displaystyle a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}\right) = ax^2 + bx + \frac{b^2}{4a}. If b^2 = 4ac, then \displaystyle \frac{b^2}{4a} = \frac{4ac}{4a} = c, so the right side is ax^2 + bx + c.
- If b is even, b^2 is a multiple of 4; if b is odd, b^2 is one more than a multiple of 4. Subtracting the multiple of 4 that is 4ac keeps that remainder, so D leaves remainder 0 or 1 on division by 4. But 2 leaves remainder 2.
- The two equations have discriminants b^2 - 4ac and b^2 - 4ca, which are the same number, so they have the same sign and the same count. (In fact the roots of the second are the reciprocals of the roots of the first.)
- In standard form the equation is x^2 - (p + q)x + pq - 1 = 0, so D = (p + q)^2 - 4(pq - 1), which is p^2 - 2pq + q^2 + 4 = (p - q)^2 + 4. A square is at least 0, so D \ge 4 > 0, and there are always two real roots.
- Multiplying by -1 leaves b^2 unchanged and changes the signs of both a and c, so ac is unchanged too. Multiplying by 3 multiplies b^2 by 9 and ac by 9, so D becomes 9D.
- No. The roots are \displaystyle \frac{-b \pm \sqrt{D}}{2a}. If \displaystyle \sqrt{D} is a whole number, both roots are rational; if it is irrational, both roots are irrational; if D < 0, neither is real. The two roots share the same \displaystyle \sqrt{D}, so they are always of the same kind. With D = 0 the single root is \displaystyle -\frac{b}{2a}, a fraction of two integers, which is rational.
- The new discriminant is b^2 - 4a(c + 1) = D - 4a, which is 4a less than before because a > 0. The gap \displaystyle \frac{\sqrt{D}}{a} between the intercepts shrinks, so they move toward each other; when D reaches 0 they meet at the vertex, and once D is negative they are gone.
- The vertex height is \displaystyle k = -\frac{D}{4a}, so D = 0 puts the vertex on the axis. Every other point of the parabola is on the side the parabola opens toward, so it is off the axis. Only the vertex is on it.
- 3 + 4i. With real coefficients, non-real solutions are \displaystyle -\frac{b}{2a} \pm \frac{\sqrt{-D}}{2a}i, a pair that differs only in the sign of the imaginary part, so 3 - 4i comes with 3 + 4i. Their shared real part, 3, is \displaystyle -\frac{b}{2a}, the x-coordinate of the vertex. One such equation is x^2 - 6x + 25 = 0.
- \displaystyle \sqrt{-4} \cdot \sqrt{-9} = 2i \cdot 3i = 6i^2 = -6. That is not equal to \displaystyle \sqrt{(-4)(-9)} = \sqrt{36} = 6. The rule \displaystyle \sqrt{a}\sqrt{b} = \sqrt{ab} is true for a and b not negative, and here both are negative, so it does not apply.