Algebra 1

The discriminant and the number of real solutions

Before you solve a quadratic equation, can you tell how many real solutions it has, and whether its parabola meets the x-axis at all?

You need

Using the quadratic formulaAlgebra 1 Roots, zeros and x-interceptsAlgebra 1

Unlocks

The discriminantAlgebra 2

Where you are 4 in The quadratic formula

Algebra9

Quadratics12

The quadratic formula4

Deriving the quadratic formula Using the quadratic formula

The discriminant and the number of real solutions

The discriminant

Why this exists

The quadratic formula solves every quadratic equation, but often a question asks for less than the full solution. Does this equation have a solution at all? Does a curve ever reach a given height? Will this trinomial factor? Each is really a question about how many solutions there are, and what kind, and one number answers it in a single line of arithmetic.

That number is the part of the formula under the square root. This lesson gives it a name, shows why it counts the solutions, and shows the same count in the graph: how many times the parabola y = ax^2 + bx + c meets the x-axis.

Definition

For a quadratic equation in standard form

ax^2 + bx + c = 0 \qquad (a \neq 0)

the discriminant is the number

D = b^2 - 4ac

It is exactly what sits under the root in the quadratic formula:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Standard form comes first: every term on one side, 0 on the other, and only then are a, b and c read off, each with its sign. The parabola y = ax^2 + bx + c has the same discriminant, and its x-intercepts are the real solutions of the equation (Roots, zeros and x-intercepts).

When the two values given by \pm are the same number, the equation has one repeated root (also called a double root). For example (x - 5)^2 = 0 has the single solution 5, produced by both factors.

Intuition

Split the formula into a middle point and a step:

x = -\frac{b}{2a} \pm \frac{\sqrt{D}}{2a}

The solutions sit one equal step either side of the middle, \displaystyle -\frac{b}{2a}. If D > 0 the step is a real, nonzero length, and there are two solutions. If D = 0 the step is zero, and both land on the same point. If D < 0 no real number squares to D, so there is no step to take and no real solution.

The graph tells the same story. Figure 1 shows y = x^2 - 4x + c for three values of c. Changing c only slides the parabola up or down.

Figure 1One parabola lifted twice, as c goes from 2 to 4 to 6. The discriminant 16 - 4c falls from 8 to 0 to -8.

With c = 2 the lowest point is below the axis, and both arms climb up through it: two x-intercepts. At c = 4 the lowest point rests on the axis: one. At c = 6 even the lowest point is above it: none. Meanwhile D = 16 - 4c goes 8, 0, -8. The discriminant is reporting where the vertex sits compared with the axis.

Results

Counting real solutions. For ax^2 + bx + c = 0 with a \neq 0, and the graph of y = ax^2 + bx + c:

D = b^2 - 4ac	Real solutions	The graph
positive	two	crosses the x-axis twice
zero	one, repeated	touches the axis at its vertex
negative	none	misses the axis

The height of the vertex. The vertex of y = ax^2 + bx + c is at

\left(-\frac{b}{2a},\; -\frac{D}{4a}\right)

Rational solutions. When a, b and c are integers, the solutions are rational exactly when D is a perfect square (0, 1, 4, 9, 16, and so on), and then ax^2 + bx + c factors into linear factors with integer coefficients. When D is positive but not a perfect square, the two solutions are irrational and no such factoring exists.

Why it is true

From the formula. Every solution of the equation comes out of the formula (Deriving the quadratic formula), so counting solutions means counting the values the formula gives. If D > 0, then \displaystyle \sqrt{D} is a positive real number and the two values differ by \displaystyle \frac{\sqrt{D}}{|a|} \neq 0: two solutions. If D = 0, both values equal \displaystyle -\frac{b}{2a}: one. If D < 0, \displaystyle \sqrt{D} is not a real number, so the formula gives no real value: none.

From the graph. Complete the square on the right-hand side:

\begin{aligned} y &= a\left(x^2 + \frac{b}{a}x\right) + c \\ &= a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c \\ &= a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a} \end{aligned}

In the second line, adding \displaystyle \frac{b^2}{4a^2} inside the bracket adds \displaystyle a \cdot \frac{b^2}{4a^2} = \frac{b^2}{4a} to y, so the same amount is subtracted outside. The third line writes \displaystyle -\frac{b^2}{4a} + c over one denominator. The squared term is 0 at \displaystyle x = -\frac{b}{2a}, so the vertex height is \displaystyle k = -\frac{D}{4a}.

Take a > 0. The parabola opens upward, k is its lowest value, and since 4a > 0, k has the opposite sign to D. If D > 0, the lowest point is below the axis and the arms rise without limit, so the curve crosses once on each side of the vertex. If D = 0, the vertex is on the axis and every other point is above it. If D < 0, even the lowest point is above the axis.

If a < 0 the parabola opens downward, k is its highest value, and \displaystyle k = \frac{D}{-4a} with -4a > 0, so k has the same sign as D. A positive D puts the top above the axis with the arms falling through it; zero puts it on the axis; negative puts it below. Either way the count matches the table.

Rational solutions. If D = n^2 for a whole number n, the formula gives \displaystyle x = \frac{-b \pm n}{2a}, a quotient of integers. In the other direction, if the trinomial factors with integer coefficients, its solutions are rational, so their difference \displaystyle \frac{\sqrt{D}}{|a|} is rational and so is \displaystyle \sqrt{D}. The square root of a whole number is either whole or irrational, so \displaystyle \sqrt{D} is whole and D is a perfect square. (That a perfect square D always leads to integer factors is also true; Example 3 shows it at work rather than proving it.)

Worked examples

Example 1

Counting without solving

How many real solutions does each equation have?

{\text{(a)}\;\; 3x^2 + 5x - 3 = 0} \qquad {\text{(b)}\;\; 4x^2 + 25 = 20x} \qquad {\text{(c)}\;\; x(2x - 3) = -4}

Solution

(a) This is already in standard form, with a = 3, b = 5, c = -3:

D = 5^2 - 4(3)(-3) = 25 + 36 = 61

Positive, so two real solutions. Since 61 is not a perfect square, both are irrational.

(b) The 20x is on the wrong side. Subtracting 20x from both sides gives 4x^2 - 20x + 25 = 0, so a = 4, b = -20, c = 25:

D = (-20)^2 - 4(4)(25) = 400 - 400 = 0

One repeated solution, at the middle \displaystyle x = -\frac{b}{2a} = \frac{20}{8} = \frac{5}{2}. Indeed 4x^2 - 20x + 25 = (2x - 5)^2.

D = (-3)^2 - 4(2)(4) = 9 - 32 = -23

Negative, so there are no real solutions. Nothing was solved, yet each answer is final.

Example 2

Where the vertex sits

Find the discriminant and the vertex, and say how the graph meets the x-axis.

{\text{(a)}\;\; y = 2x^2 + 8x + 5} \qquad {\text{(b)}\;\; y = -3x^2 + 6x - 4}

Solution

(a) D = 8^2 - 4(2)(5) = 64 - 40 = 24. The vertex is at \displaystyle x = -\frac{8}{2(2)} = -2, with height \displaystyle k = -\frac{D}{4a} = -\frac{24}{8} = -3. Substituting checks it: 2(4) - 16 + 5 = -3. The parabola opens upward (a = 2 > 0) from a lowest point 3 below the axis, so it crosses twice, as D > 0 says. The formula places the crossings at

x = \frac{-8 \pm \sqrt{24}}{4} = -2 \pm \frac{\sqrt{6}}{2}

(b) D = 6^2 - 4(-3)(-4) = 36 - 48 = -12. The vertex is at \displaystyle x = -\frac{6}{2(-3)} = 1, with height \displaystyle k = -\frac{-12}{4(-3)} = -1, and indeed -3 + 6 - 4 = -1. The parabola opens downward from a highest point already below the axis, so it never meets the axis, and -3x^2 + 6x - 4 = 0 has no real solutions.

Example 3

Will it factor?

Decide whether each trinomial factors with integer coefficients, and factor it if it does.

{\text{(a)}\;\; 10x^2 - 7x - 12} \qquad {\text{(b)}\;\; 10x^2 - 7x - 11}

Solution

With a = 10 there are many pairs to try, so the discriminant settles the question first, the same test used in Using the quadratic formula.

(a) The discriminant is

D = (-7)^2 - 4(10)(-12) = 49 + 480 = 529 = 23^2

a perfect square, so it factors. The formula gives \displaystyle x = \frac{7 \pm 23}{20}, that is \displaystyle \frac{30}{20} = \frac{3}{2} and \displaystyle -\frac{16}{20} = -\frac{4}{5}. The solution \displaystyle \frac{3}{2} comes from the factor 2x - 3, and \displaystyle -\frac{4}{5} from 5x + 4. Multiplying out confirms it:

\begin{aligned} (2x - 3)(5x + 4) &= 10x^2 + 8x - 15x - 12 \\ &= 10x^2 - 7x - 12 \end{aligned}

(b) D = 49 + 440 = 489. Since 22^2 = 484 and 23^2 = 529, it is not a perfect square, so the trinomial does not factor with integer coefficients. The equation 10x^2 - 7x - 11 = 0 still has two real solutions, \displaystyle \frac{7 \pm \sqrt{489}}{20}; they are irrational.

Example 4

Choosing the constant

For which values of k does x^2 - 10x + k = 0 have (a) one real solution, (b) two, (c) none?

Solution

Here a = 1, b = -10, c = k, so D = 100 - 4k, and each case is a condition on D.

(a) One solution means 100 - 4k = 0, so k = 25. Then x^2 - 10x + 25 = (x - 5)^2, and the one solution is x = 5.

(b) Two means 100 - 4k > 0, so 4k < 100 and k < 25.

This is Figure 1 again: y = x^2 - 10x + k has its vertex at x = 5, at height 25 - 50 + k = k - 25, so k slides it up and down, and k = 25 is the one height at which the vertex rests on the axis.

Example 5

Reading the sign from a graph

(a) Figure 2 shows three parabolas. Give the sign of each one’s discriminant.

Figure 2Three parabolas, no equations. The number of x-intercepts gives the sign of each discriminant.

(b) A parabola opens downward and its vertex is (4, 3). Is its discriminant positive, zero or negative?

Solution

(a) The number of x-intercepts is the sign. A crosses twice, so its discriminant is positive. B touches at its vertex, so zero. C misses, so negative. No equation is needed.

(b) From \displaystyle k = -\frac{D}{4a}, D = -4ak = -4a(3) = -12a. Opening downward means a < 0, so -12a is positive. The picture agrees: the highest point is above the axis and the arms fall through it, giving two x-intercepts.

Example 6

Can the truck pass under the arch?

The inside curve of a stone arch is modeled by \displaystyle h = -\frac{1}{4}x^2 + 2x, where x is the distance in meters from one foot of the arch and h is the height in meters. Dragan’s truck is 3.5 meters tall and Ezio’s is 4.5 meters tall. Does the arch reach each height, and over what width?

Solution

The arch reaches 3.5 meters if \displaystyle -\frac{1}{4}x^2 + 2x = 3.5 has a real solution. Multiplying both sides by -4, which does not change the solutions since -4 \neq 0, and moving everything to one side gives

x^2 - 8x + 14 = 0, \qquad D = 64 - 56 = 8

Positive, so the arch is exactly 3.5 meters high at two places, \displaystyle x = \frac{8 \pm \sqrt{8}}{2} = 4 \pm \sqrt{2}. The arch opens downward, so between those places it is higher. Dragan has a stretch \displaystyle 2\sqrt{2} \approx 2.83 meters wide with at least 3.5 meters of headroom, centered on the middle of the arch, and his truck passes if it is narrower than that.

Figure 3The arch reaches 3.5 meters at two places and is higher between them. It never reaches 4.5 meters.

For 4.5 meters the same steps give

x^2 - 8x + 18 = 0, \qquad D = 64 - 72 = -8

Negative: the arch is never 4.5 meters high, and Ezio’s truck cannot pass anywhere. The top lies between the two: for a height H the equation is x^2 - 8x + 4H = 0, with D = 64 - 16H, which is 0 only for H = 4. So the arch is 4 meters tall, reached once, at x = 4.

Boundaries

Real numbers only. A negative discriminant means no real solutions, and a graph with no x-intercepts. Over the complex numbers such an equation does have two solutions; that is taken up in The discriminant.

It needs a \neq 0. With a = 0 the equation is linear and b^2 - 4ac = b^2 says nothing: 2x + 3 = 0 would give D = 4, yet it has exactly one solution.

It answers = 0, in standard form. The discriminant of ax^2 + bx + c is about ax^2 + bx + c = 0. To ask whether the expression ever equals another number, move that number across first, as in Example 6.

The perfect-square test needs integer coefficients. With fractions, multiply through by a common denominator first.

The sign counts; it does not locate. D alone does not say where the solutions are or which way the parabola opens.

Where this goes next

Setting a quadratic equal to a target and counting, as with the arch, is the same move used in Systems with a curve to decide how often a line meets a parabola. The vertex height \displaystyle -\frac{D}{4a} feeds straight into graphing from standard form. Whether the graph crosses the axis also settles the sign of ax^2 + bx + c everywhere, which is the start of Quadratic inequalities.

Exercises

How many real solutions?

Write each equation in standard form if needed, find the discriminant, and say how many real solutions it has.

x^2 + 13x + 30 = 0
16x^2 - 24x + 9 = 0
3x^2 = 4x + 2
x(x - 8) = -16
(x - 2)^2 = -3x

The graph

For each graph, find the discriminant and the vertex, and say how it meets the x-axis.

y = 3x^2 + 12x + 7
y = -x^2 - 6x - 11
y = -3x^2 + 12x - 12
A parabola opens upward and its lowest point is (-1, 2). Is its discriminant positive, zero or negative, and how many real solutions does the matching equation have?
A parabola has its vertex at (2, 6) and passes through (0, 2). Find the sign of its discriminant.

Choosing a number

For what value of k does x^2 + 14x + k = 0 have exactly one solution, and what is it?
For which values of k does 3x^2 - 6x + k = 0 have two real solutions?
For which values of b does 4x^2 + bx + 1 = 0 have no real solutions?
Which of 12x^2 + x - 6 and 12x^2 + x - 5 factors with integer coefficients? Factor it.

Targets

A toll road’s revenue is modeled by R = -40p^2 + 480p, where p is the toll in dollars and R is the revenue in dollars per hour. (a) Can the revenue reach 1,600 dollars an hour? (b) 1,400 dollars an hour? (c) Which revenue is reached at exactly one toll?
Can x^2 + 8x + 20 ever equal 2? Can it ever equal 5?

Reasoning

Explain, from the formula, why when D = 0 the one solution is the x-coordinate of the vertex.
Catalina says: “If ax^2 + bx + c = 0 has no real solutions, then neither does -ax^2 - bx - c = 0.” Is she right? Explain with the discriminant and with the graph.

Answers

D = 169 - 120 = 49, two solutions. A perfect square, so they are rational: x = -3 and x = -10.
D = 576 - 576 = 0, one solution, \displaystyle x = \frac{3}{4}. The left side is (4x - 3)^2.
Standard form 3x^2 - 4x - 2 = 0: D = 16 + 24 = 40, two solutions.
Standard form x^2 - 8x + 16 = 0: D = 64 - 64 = 0, one solution, x = 4.
Expanding, x^2 - 4x + 4 = -3x, so x^2 - x + 4 = 0: D = 1 - 16 = -15, no real solutions.
D = 144 - 84 = 60, vertex (-2, -5): crosses twice. Check: \displaystyle k = -\frac{60}{12} = -5.
D = 36 - 44 = -8, vertex (-3, -2): misses. It opens downward from a top below the axis.
D = 144 - 144 = 0, vertex (2, 0): touches the axis there. It is -3(x - 2)^2.
Negative, no real solutions. D = -4ak = -4a(2) = -8a, negative since a > 0: the lowest point is above the axis.
Positive. Writing y = a(x - 2)^2 + 6 and substituting (0, 2) gives 4a + 6 = 2, so a = -1 and y = -x^2 + 4x + 2, with D = 16 + 8 = 24. The top is above the axis and the parabola opens downward.
k = 49, and the solution is x = -7. D = 196 - 4k = 0, and then the left side is (x + 7)^2.
k < 3. D = 36 - 12k > 0.
-4 < b < 4. D = b^2 - 16 < 0, so b^2 < 16.
12x^2 + x - 6 = (4x + 3)(3x - 2): its D = 1 + 288 = 289 = 17^2. The other has D = 1 + 240 = 241, between 15^2 and 16^2, so it does not factor that way.
(a) No. -40p^2 + 480p = 1600 divides by -40 to p^2 - 12p + 40 = 0, with D = 144 - 160 = -16. (b) Yes: p^2 - 12p + 35 = 0 has D = 144 - 140 = 4, at p = 5 or p = 7. (c) R = 1440, at p = 6: \displaystyle p^2 - 12p + \frac{R}{40} = 0 has \displaystyle D = 144 - \frac{R}{10}, which is 0 when R = 1440. That is the greatest revenue.
Not 2: x^2 + 8x + 18 = 0 has D = 64 - 72 = -8. Yes 5: x^2 + 8x + 15 = 0 has D = 64 - 60 = 4, at x = -3 and x = -5.
With D = 0 the formula gives \displaystyle x = \frac{-b \pm 0}{2a} = -\frac{b}{2a}, which is the x-coordinate of the vertex. The vertex height is then \displaystyle -\frac{0}{4a} = 0, so the vertex is the point where the graph touches the axis.
Yes. The second discriminant is (-b)^2 - 4(-a)(-c) = b^2 - 4ac, the same number, so if one is negative so is the other. On the graph, y = -ax^2 - bx - c is y = ax^2 + bx + c flipped over the x-axis, and a flip over the axis keeps every point on the axis where it was and moves no point onto it.