Algebra 1 · A.REI.B.4 · A.REI.B.4B

Using the quadratic formula

The quadratic formula solves every quadratic. How do you put numbers into it without losing a sign, and how do you tell when another method gets there faster?

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Deriving the quadratic formulaAlgebra 1 Solving quadratics by factoringAlgebra 1 Simplifying radicalsGrade 9

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Maximum, minimum and projectile problemsAlgebra 1 The discriminant and the number of real solutionsAlgebra 1

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

When you reach for it

Completing the square solves every quadratic equation, but it makes you do the same algebra every time. Deriving the quadratic formula does that algebra once, with the letters a, b and c in place of numbers, and hands back the finished answer. The equation

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

has the solutions

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

So using the formula is not new algebra. It is substitution and arithmetic, and the whole skill is doing that arithmetic without dropping a sign or canceling something that cannot be canceled.

Reach for it when a quadratic does not factor over the integers, when its coefficients are large or awkward, when you want a decimal answer, or when you are not sure whether it factors and do not want to hunt. It never fails: if the equation has solutions, the formula finds them, and if it has none that are real, the formula tells you so. When a quadratic factors quickly, factoring is faster, and when there is no x term, taking square roots is. The part of the procedure headed “Choosing how to solve a quadratic” says how to tell, in a few seconds, which road is shortest.

The procedure

Substituting into the formula

Take the equation 2x^2 = 3x + 5.

Put the equation in standard form. The formula was derived from ax^2 + bx + c = 0, with everything on one side and zero on the other. The letters a, b and c mean the coefficients of that equation and of no other arrangement. Here the 5 is not yet c. Subtract 3x and 5 from both sides, which keeps the equation true:

2x^2 - 3x - 5 = 0

Read off a, b and c, each with its sign. The expression 2x^2 - 3x - 5 means 2x^2 + (-3)x + (-5), so the minus signs belong to the numbers:

a = 2, \qquad b = -3, \qquad c = -5

A term that is missing has coefficient 0. In x^2 - 7 = 0, b = 0.

Work out b^2 - 4ac on its own first. This is where most of the arithmetic is, so it pays to do it in one place. Put each number in brackets, so that a negative stays negative while it is squared and multiplied:

\begin{aligned} b^2 - 4ac &= (-3)^2 - 4(2)(-5) \\ &= 9 + 40 = 49 \end{aligned}

The square of -3 is +9, and -4 times 2 times -5 is +40, because two negatives multiply to a positive. This number has a name, the discriminant, and its sign tells you what kind of answer is coming before you finish.

Substitute into the formula. The -b at the front means the opposite of b, so with b = -3 it is -(-3) = 3:

\begin{aligned} x &= \frac{-(-3) \pm \sqrt{49}}{2(2)} \\[6pt] &= \frac{3 \pm 7}{4} \end{aligned}

Split the \pm into two answers. One solution uses the plus, the other the minus:

\begin{aligned} x &= \frac{3 + 7}{4} = \frac{5}{2} \\[6pt] x &= \frac{3 - 7}{4} = -1 \end{aligned}

Check. Both work:

\begin{aligned} x = \tfrac{5}{2}: &\quad 2 \cdot \frac{25}{4} - \frac{15}{2} - 5 = 0 \\[6pt] x = -1: &\quad 2 + 3 - 5 = 0 \end{aligned}

When the square root does not come out whole, simplify the radical, then look for a factor shared by every term of the numerator and by the denominator. The fraction bar groups the whole numerator, so the denominator divides both -b and the square root. A factor that divides only one of them cannot be canceled. Example 2 shows this.

What the formula says on the graph

It helps to see what the formula is saying. Split the fraction into its two parts:

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

The first part is the same for both solutions. The second is added once and subtracted once. So the two solutions sit an equal distance either side of \displaystyle -\frac{b}{2a}, which is where the axis of symmetry of the parabola y = ax^2 + bx + c crosses the x-axis.

Figure 1The two solutions sit an equal distance either side of the axis of symmetry, and that distance is the square-root part of the formula.

For 2x^2 - 3x - 5 = 0 the center is \displaystyle \frac{3}{4} and the distance either side is \displaystyle \frac{7}{4}, and indeed

\begin{aligned} \frac{3}{4} + \frac{7}{4} &= \frac{5}{2} \\[6pt] \frac{3}{4} - \frac{7}{4} &= -1 \end{aligned}

When b^2 - 4ac = 0 the distance shrinks to nothing and the two solutions become one. When b^2 - 4ac is negative there is no real square root, and the parabola never reaches the axis.

Choosing how to solve a quadratic

There are four ways to solve a quadratic equation: factoring, the square root property, completing the square, and the formula. Done correctly, they always agree. A quadratic has the solutions it has, and no method can add one or lose one.

What they do not agree on is effort. Solve x^2 = 49 with the formula and you write out a = 1, b = 0, c = -49, compute 0 + 196, take its root and divide by 2, all to reach x = \pm 7, which you could have seen at a glance. Try to factor 3x^2 - 5x - 1 and you can test factor pairs all afternoon, because it does not factor at all.

So choosing is not a fifth method. It is looking at the equation for a few seconds before you start, and picking the method that reaches its solutions in the fewest safe steps. Each method has one thing it needs from the equation.

The square root property needs the variable in one place. It solves ax^2 = k and a(x - h)^2 = k directly: get the square alone, then take the root of both sides with a \pm. The sign to look for is a quadratic with no x term, or a squared bracket with nothing else containing x.

Factoring needs zero on one side and an expression that splits into two factors, after which the zero product property turns one equation into two small ones. It is always possible, and always quick, when there is no constant term, since then x is a common factor. Otherwise it works exactly when the trinomial factors over the integers, and the discriminant tells you in advance whether it does.

Completing the square always works. It is the natural choice when the coefficient of x^2 is 1 and the coefficient of x is even, because then half of it is a whole number and no fractions appear.

The quadratic formula always works, for every equation ax^2 + bx + c = 0 with a \neq 0. It is what you use when nothing about the equation offers a shortcut.

You can see the difference between them on a graph. The solutions of ax^2 + bx + c = 0 are the places where the parabola y = ax^2 + bx + c crosses the x-axis.

Figure 2Where the roots land decides the method. Whole numbers, factor. Mirror images across the y-axis, square roots. Anywhere else, the formula.

At the top, x^2 - x - 6 = 0 crosses at -2 and 3, exactly on the ticks. Whole-number (or fraction) solutions are what factoring finds, because (x + 2)(x - 3) is those two numbers written as factors.

In the middle, x^2 - 5 = 0 has no x term. Every parabola is symmetric about its own axis, and with no x term that axis is the y-axis itself, so the two solutions are the same number with opposite signs, \displaystyle -\sqrt{5} and \displaystyle \sqrt{5}. That is exactly the shape of an answer \displaystyle \pm\sqrt{k}, which is what the square root property produces.

At the bottom, x^2 - 2x - 4 = 0 crosses between the ticks and off center, at \displaystyle 1 \pm \sqrt{5}. Factoring over the integers can never find numbers like that, and the square root property cannot start until the equation is rewritten. The formula, or completing the square, is the tool that reaches them.

So the choice is really a guess about what the answer looks like. Look for the cheap shapes first, since they are cheap when they occur and the formula is always waiting behind them.

Four facts that make the choice quick

Every method gives the same solutions. The solutions belong to the equation, not to the method. If two methods disagree, one of them was done wrongly.

The factoring test. When a, b and c are integers, ax^2 + bx + c factors into two linear factors with integer coefficients exactly when its discriminant b^2 - 4ac is a perfect square: 0, 1, 4, 9, 16, 25, \dots

No middle term, no factoring needed. When b = 0, the equation is ax^2 = -c, and the square root property gives

x = \pm\sqrt{-\frac{c}{a}}

which are real numbers when \displaystyle -\frac{c}{a} \ge 0, and a complex pair otherwise.

No constant term, always factor. When c = 0, the equation is x(ax + b) = 0, and its solutions are x = 0 and \displaystyle x = -\frac{b}{a}.

Why the methods agree, and why the factoring test works

Why the methods agree. Each method is a chain of rewrites, and each rewrite is a move that keeps the same solutions: adding the same thing to both sides, multiplying both sides by a number that is not zero, factoring or expanding an expression, taking the square root with a \pm. A chain of moves that each keep the solutions keeps them all the way to the end. The formula is not an exception. It is completing the square done once on the letters a, b and c, so it is one particular chain of those same moves.

Why the factoring test works. Suppose first that the quadratic factors as (px + q)(rx + s) with integers p, q, r, s. Multiplying out,

(px + q)(rx + s) = pr\,x^2 + (ps + qr)\,x + qs

so a = pr, b = ps + qr and c = qs. Then

\begin{aligned} b^2 - 4ac &= (ps + qr)^2 - 4prqs \\ &= p^2s^2 + 2pqrs + q^2r^2 - 4pqrs \\ &= p^2s^2 - 2pqrs + q^2r^2 \\ &= (ps - qr)^2 \end{aligned}

which is the square of the integer ps - qr.

Now suppose instead that b^2 - 4ac = m^2 for a whole number m. The formula gives the solutions \displaystyle \frac{-b + m}{2a} and \displaystyle \frac{-b - m}{2a}, and both are fractions of integers. A quadratic with roots r_1 and r_2 equals a(x - r_1)(x - r_2), so it factors with rational coefficients, and the denominators can always be cleared by sharing the factor a out between the two brackets. (That last step is a small theorem of its own, Gauss’s lemma.) Either way round, a perfect square discriminant and factoring over the integers go together.

Worked examples

Example 1

A constant on the wrong side

Solve x^2 = 4x + 21.

Solution

Move everything to the left so the right side is zero:

x^2 - 4x - 21 = 0

So a = 1, b = -4 and c = -21. The number under the root is

\begin{aligned} b^2 - 4ac &= (-4)^2 - 4(1)(-21) \\ &= 16 + 84 = 100 \end{aligned}

and \displaystyle \sqrt{100} = 10. Substituting, with -b = 4:

x = \frac{4 \pm 10}{2(1)} = \frac{4 \pm 10}{2}

\begin{aligned} x &= \frac{14}{2} = 7 \\[6pt] x &= \frac{-6}{2} = -3 \end{aligned}

Check. 7^2 = 49 and 4(7) + 21 = 49. (-3)^2 = 9 and 4(-3) + 21 = 9.

This one also factors, as (x - 7)(x + 3) = 0, and the formula agrees with factoring, as it must. A whole-number square root under the formula is the sign that factoring would have worked, which is the factoring test at work.

Example 2

Simplifying the radical and the fraction

Solve 3x^2 + 6x - 2 = 0.

Solution

Here a = 3, b = 6 and c = -2.

\begin{aligned} b^2 - 4ac &= 6^2 - 4(3)(-2) \\ &= 36 + 24 = 60 \end{aligned}

Since 60 = 4 \cdot 15 and 4 is a perfect square, \displaystyle \sqrt{60} = \sqrt{4}\sqrt{15} = 2\sqrt{15}.

\begin{aligned} x &= \frac{-6 \pm 2\sqrt{15}}{2(3)} \\[6pt] &= \frac{-6 \pm 2\sqrt{15}}{6} \end{aligned}

Every term in the numerator, and the denominator, is divisible by 2. Take the 2 out of the numerator as a common factor and cancel it against the 6:

\begin{aligned} x &= \frac{2\left(-3 \pm \sqrt{15}\right)}{2 \cdot 3} \\[6pt] &= \frac{-3 \pm \sqrt{15}}{3} \end{aligned}

The canceling is allowed only because the 2 is a factor of the whole numerator. As decimals the solutions are about 0.291 and -2.291.

Example 3

Fractions in the equation

Solve \displaystyle \frac{1}{3}x^2 - \frac{1}{2}x - 1 = 0.

Solution

You could substitute \displaystyle a = \frac{1}{3} and \displaystyle b = -\frac{1}{2} directly, but the arithmetic is kinder with whole numbers. Multiply both sides by 6, the lowest common denominator. Multiplying both sides of an equation by the same non-zero number does not change its solutions, and the right side stays 0:

2x^2 - 3x - 6 = 0

Now a = 2, b = -3, c = -6:

\begin{aligned} b^2 - 4ac &= (-3)^2 - 4(2)(-6) \\ &= 9 + 48 = 57 \end{aligned}

x = \frac{3 \pm \sqrt{57}}{4}

Since 57 = 3 \cdot 19 has no square factor, \displaystyle \sqrt{57} is already simplified. The two terms of the numerator, 3 and \displaystyle 1 \cdot \sqrt{57}, share no whole-number factor with each other and the 4, so nothing cancels and this is the final form.

Example 4

Zero, or a negative, under the root

Solve (a) 4x^2 - 12x + 9 = 0 and (b) 2x^2 - 4x + 5 = 0.

Solution

(a) Here a = 4, b = -12, c = 9:

\begin{aligned} b^2 - 4ac &= 144 - 4(4)(9) \\ &= 144 - 144 = 0 \end{aligned}

x = \frac{12 \pm 0}{8} = \frac{3}{2}

Adding and subtracting zero give the same number, so there is exactly one solution. The left side was a perfect square all along, (2x - 3)^2, and on the graph the vertex of the parabola sits right on the x-axis.

(b) Here a = 2, b = -4, c = 5:

\begin{aligned} b^2 - 4ac &= 16 - 4(2)(5) \\ &= 16 - 40 = -24 \end{aligned}

No real number squares to -24, so the equation has no real solutions. That is a finished answer, not a failure: the parabola y = 2x^2 - 4x + 5 lies wholly above the x-axis. Over the complex numbers, \displaystyle \sqrt{-24} = 2i\sqrt{6} and the formula gives

x = \frac{4 \pm 2i\sqrt{6}}{4} = 1 \pm \frac{\sqrt{6}}{2}\,i

which is the business of complex roots.

Example 5

Expanding first

Solve (2x + 1)(x - 3) = x - 5.

Solution

The right side is not zero, so the two factors on the left tell you nothing on their own. Multiply out and collect everything on one side:

\begin{aligned} 2x^2 - 5x - 3 &= x - 5 \\ 2x^2 - 6x + 2 &= 0 \end{aligned}

Every coefficient is even. Dividing both sides by 2 keeps the same solutions and makes the numbers smaller:

x^2 - 3x + 1 = 0

So a = 1, b = -3, c = 1, and b^2 - 4ac = 9 - 4 = 5:

x = \frac{3 \pm \sqrt{5}}{2}

Using 2x^2 - 6x + 2 = 0 instead gives

\begin{aligned} x &= \frac{6 \pm \sqrt{20}}{4} = \frac{6 \pm 2\sqrt{5}}{4} \\[6pt] &= \frac{3 \pm \sqrt{5}}{2} \end{aligned}

the same thing. Dividing out the common factor first just saves the canceling at the end, and it is worth doing whatever method follows.

Example 6

Reading three equations

Choose a method for each equation, then solve it.

(a) 3x^2 = 48

(b) x^2 + 7x + 12 = 0

Solution

(a) There is no x term, so the variable appears once. That is the square root property’s case. Divide by 3 to get the square alone, then root both sides, keeping both signs:

x^2 = 16 \qquad\Longrightarrow\qquad x = \pm 4

(b) Zero is already on one side and the coefficient of x^2 is 1, so look for two numbers that multiply to 12 and add to 7. They are 3 and 4:

(x + 3)(x + 4) = 0

so x = -3 or x = -4.

(c) Factor pairs for 2 \cdot (-7) that add to -3 do not come to mind, and the discriminant says why: (-3)^2 - 4(2)(-7) = 9 + 56 = 65, which is not a perfect square, so this trinomial does not factor over the integers. Use the formula with a = 2, b = -3, c = -7:

x = \frac{3 \pm \sqrt{65}}{4}

Example 7

A missing middle term

Solve 3t^2 - 10 = 0.

Solution

The formula does not care what the variable is called. There is no t term, so b = 0, with a = 3 and c = -10:

\begin{aligned} b^2 - 4ac &= 0^2 - 4(3)(-10) \\ &= 120 \end{aligned}

and \displaystyle \sqrt{120} = \sqrt{4 \cdot 30} = 2\sqrt{30}. With -b = 0:

t = \frac{0 \pm 2\sqrt{30}}{6} = \pm\frac{\sqrt{30}}{3}

Taking square roots gets there faster. Divide by 3 and take both roots, then make the denominator rational:

\begin{aligned} t^2 &= \frac{10}{3} \\[6pt] t &= \pm\sqrt{\frac{10}{3}} = \pm\frac{\sqrt{10}}{\sqrt{3}} \\[6pt] &= \pm\frac{\sqrt{30}}{3} \end{aligned}

The formula still works, which is the point; it is simply not the shortest road when b = 0.

Example 8

A squared bracket already in place

Solve 2(x - 3)^2 = 50.

Solution

The only place x appears is inside the bracket, so this is the square root property’s case with x - 3 in the role of the variable. Divide by 2, then root:

\begin{aligned} (x - 3)^2 &= 25 \\ x - 3 &= \pm 5 \\ x &= 3 \pm 5 \end{aligned}

so x = 8 or x = -2.

Expanding first would also work. It gives 2x^2 - 12x + 18 = 50, then x^2 - 6x - 16 = 0, then (x - 8)(x + 2) = 0: the same two answers, after three extra lines that undo the work the equation had already done for you. When an equation arrives already in the form a(x - h)^2 = k, leave the bracket alone.

Example 9

No constant term

Solve 5x^2 = 20x.

Solution

There is no constant term, so once everything is on one side, x is a common factor. Subtract 20x from both sides so that zero stands alone:

\begin{aligned} 5x^2 - 20x &= 0 \\ 5x(x - 4) &= 0 \end{aligned}

A product is zero only when a factor is, so x = 0 or x = 4.

It is tempting to divide both sides of 5x^2 = 20x by 5x and read off x = 4. That division is allowed only when 5x \neq 0, so it silently assumes x \neq 0, and x = 0 is a solution: 5(0)^2 = 0 = 20(0). Dividing by an expression containing the variable can throw a solution away. Factoring keeps it.

Example 10

Letting the discriminant decide

Decide whether each equation can be solved by factoring, then solve it.

(a) 6x^2 + x - 15 = 0

(b) 6x^2 + x - 14 = 0

Solution

With a = 6, trial and error has many pairs to test, so compute the discriminant first. It takes one line and settles the question.

(a) The discriminant is

\begin{aligned} b^2 - 4ac &= 1 - 4(6)(-15) \\ &= 1 + 360 = 361 = 19^2 \end{aligned}

a perfect square, so the trinomial factors. The two numbers that multiply to 6 \cdot (-15) = -90 and add to 1 are 10 and -9:

\begin{aligned} 6x^2 + 10x - 9x - 15 &= 0 \\ 2x(3x + 5) - 3(3x + 5) &= 0 \\ (2x - 3)(3x + 5) &= 0 \end{aligned}

so \displaystyle x = \frac{3}{2} or \displaystyle x = -\frac{5}{3}.

(b) b^2 - 4ac = 1 + 336 = 337. Since 18^2 = 324 and 19^2 = 361, 337 is not a perfect square, so there is no factoring to find. The formula gives

x = \frac{-1 \pm \sqrt{337}}{12}

The two equations differ by 1 in one coefficient, and one has fraction solutions while the other has irrational ones. Looking at the coefficients cannot tell the two apart; the discriminant can.

Example 11

Completing the square, when it is the short way

Solve x^2 + 10x - 3 = 0.

Solution

The discriminant is 100 + 12 = 112, not a perfect square, so factoring is out. The coefficient of x^2 is 1 and the coefficient of x is even, which is exactly when completing the square is at its shortest. Move the constant across, then add \displaystyle \left(\frac{10}{2}\right)^2 = 25 to both sides:

\begin{aligned} x^2 + 10x &= 3 \\ x^2 + 10x + 25 &= 28 \\ (x + 5)^2 &= 28 \\ x + 5 &= \pm\sqrt{28} = \pm 2\sqrt{7} \\ x &= -5 \pm 2\sqrt{7} \end{aligned}

Check with the formula: \displaystyle x = \frac{-10 \pm \sqrt{112}}{2} = \frac{-10 \pm 4\sqrt{7}}{2} = -5 \pm 2\sqrt{7}. The same answer, as it must be, but the formula needed a larger radical simplified and a fraction reduced.

Example 12

The golden rectangle

A rectangle is 1 unit tall and x units long. Cutting a 1 by 1 square off one end leaves a smaller rectangle, 1 by x - 1, with exactly the same shape: its long side is to its short side as x is to 1. Find x.

Figure 3Cut the square off and the shaded piece has the same shape as the whole rectangle.

Solution

In the small rectangle the long side is 1 and the short side is x - 1. The same shape means the same ratio of long side to short side:

\frac{x}{1} = \frac{1}{x - 1}

Multiply both sides by x - 1 to clear the fraction, then move everything to one side:

\begin{aligned} x(x - 1) &= 1 \\ x^2 - x - 1 &= 0 \end{aligned}

Here a = 1, b = -1, c = -1, and b^2 - 4ac = 1 + 4 = 5, not a perfect square, so this is a job for the formula:

x = \frac{1 \pm \sqrt{5}}{2}

The equation has two solutions, but a length cannot be negative, and the minus sign gives about -0.618. So

x = \frac{1 + \sqrt{5}}{2} \approx 1.618

This number is the golden ratio. The formula solved the equation; deciding which solution answers the question was a separate step, and it always is when the equation comes from a situation.

Boundaries

a must not be 0. With a = 0 the equation bx + c = 0 is linear, and the formula would divide by 2a = 0. The formula is for quadratics only.

Standard form comes first. The formula answers ax^2 + bx + c = 0 and nothing else. Reading c off an equation whose right side is not 0 solves a different equation.

Real solutions need b^2 - 4ac \ge 0. Positive gives two real solutions, zero gives one, and negative gives none that are real. When it is negative the parabola misses the x-axis, factoring over the integers finds nothing, and the three methods that end by taking a square root all arrive at the square root of a negative number. Over the complex numbers that is still a pair of solutions; see when the formula gives complex roots.

The factoring test needs integer coefficients. It speaks about factoring over the integers, so the coefficients must be integers first. With fractions or decimals, multiply through to clear them, as in exercises 20 and 21. With an irrational coefficient, such as in \displaystyle x^2 - \sqrt{2}\,x - 4 = 0, the test says nothing and the formula is the method.

Factoring over the integers is not the only kind. Over the real numbers, any quadratic with a non-negative discriminant factors as a(x - r_1)(x - r_2), with its two roots in the brackets. Using the formula is, in that sense, factoring with the hard part done for you.

The formula solves the equation, not the problem. When the quadratic comes from a length, a time or a count, each solution has to be tested against the situation, as in Example 12.

Exact first, decimals last. Rounding \displaystyle \sqrt{57} before dividing carries the rounding error through every later step. Keep the radical until the end, then approximate once.

The choice of method is about effort, never about correctness. A slow method done right is right. The only real cost of a poor choice is more arithmetic, and so more places for a slip.

It stops at degree two. For cubic and quartic equations there are formulas too, but they are long enough that nobody reaches for them first, and from degree five on there is provably no general formula built from roots at all. Past quadratics, looking for factors first stops being a shortcut and becomes the main method.

Where this goes next

The number under the root, b^2 - 4ac, is worth computing even when you do not want the solutions: the discriminant and its sign tell you how many real solutions there are without finishing the formula, and when it is negative the formula gives complex roots.

The center \displaystyle -\frac{b}{2a} in the first figure is the x-coordinate of the vertex, so the formula also hands you the axis of the parabola for free.

Every applied quadratic ends in the choice of method. In projectile motion and other quadratic word problems, the model hands you an equation and the method is yours to pick, with the extra step of throwing out solutions the situation does not allow. Quadratic inequalities begin by solving the matching equation, so the same judgment comes first there too. The formula is also the working tool of equations that are quadratic in something else, such as \sin x, and of the eigenvalues of a 2 \times 2 matrix.

The factoring test is the quadratic case of a wider idea. For a polynomial of any degree with integer coefficients, the rational root theorem lists every fraction that could possibly be a root, which is the higher-degree way of asking whether it factors.

Exercises

Standard form

Write each equation in standard form and name a, b and c.

5x = 3 - 2x^2
x^2 = 9x
7 - 4t^2 = 0

Solve with the formula

Give exact answers, with radicals simplified.

2x^2 + x - 6 = 0
2x^2 + 4x - 3 = 0
5n^2 - 2n - 1 = 0
x^2 - x + 3 = 0

Choose a method, then solve

x^2 - 8x = 0
(x + 2)^2 = 18
3x^2 - 5x - 1 = 0
x^2 + 6x + 4 = 0
2x^2 + 7x + 3 = 0
3(x - 1)^2 - 7 = 20

Does it factor?

Use the discriminant to decide whether the equation factors over the integers, then solve it.

6x^2 - 7x - 20 = 0
4x^2 + 5x - 3 = 0
9x^2 - 12x + 4 = 0

Rearrange first

x(x + 8) = 2
(2x - 1)(x + 2) = 3x
(2x - 1)^2 = 2x^2 + 7
\displaystyle \frac{1}{2}x^2 + \frac{1}{3}x - 1 = 0
0.5x^2 + 0.4x - 0.1 = 0

In context

A rectangle is 2 m longer than it is wide, and its area is 10 m^2. Find its width exactly, and to two decimal places.
A ball’s height in meters after t seconds is h = -5t^2 + 20t + 1. At what times is it 16 m high?
A stone is dropped from a cliff 144 feet high, and its height after t seconds is h = 144 - 16t^2. When does it reach the ground?
A ball is thrown upward from 5 feet above the ground, and its height after t seconds is h = -16t^2 + 32t + 5. When does it land?
Where does the parabola y = x^2 - 4x + 1 cross the x-axis? Give exact values and values to two decimal places.

Reasoning

Explain why x^2 + 3x = 4 cannot be solved by substituting a = 1, b = 3, c = 4 into the formula.
Show that the two solutions the formula gives always add to \displaystyle -\frac{b}{a}.
Explain why, when a and c have opposite signs, the equation ax^2 + bx + c = 0 always has two real solutions.
Show that \displaystyle \frac{-6 \pm 2\sqrt{15}}{6} equals \displaystyle -1 \pm \frac{\sqrt{15}}{3}, and explain why it does not equal \displaystyle -1 \pm 2\sqrt{15}.
Two students solve the same quadratic equation, one by factoring and one with the formula. Explain why, if both work correctly, they must get the same solutions.
Solve x^2 = 5x by factoring. Then explain why dividing both sides by x finds only one of the two solutions.
Show that when c = 0 and a, b are integers, the discriminant of ax^2 + bx + c is a perfect square, which agrees with the fact that such an equation always factors.
Explain why completing the square on x^2 + bx + c = 0 involves no fractions when b is even, and use it to show that the solutions are then \displaystyle x = -\frac{b}{2} \pm \sqrt{\frac{b^2}{4} - c}.

Answers

2x^2 + 5x - 3 = 0, so a = 2, b = 5, c = -3. Add 2x^2 to both sides and subtract 3.
x^2 - 9x = 0, so a = 1, b = -9, c = 0. Subtract 9x; there is no constant term.
4t^2 - 7 = 0, so a = 4, b = 0, c = -7. Multiply both sides by -1 and reorder. Keeping -4t^2 + 7 = 0, with a = -4, b = 0, c = 7, is also correct and gives the same solutions.
\displaystyle x = \frac{3}{2} or x = -2. b^2 - 4ac = 1 + 48 = 49, so \displaystyle x = \frac{-1 \pm 7}{4}.
\displaystyle x = \frac{-2 \pm \sqrt{10}}{2}. b^2 - 4ac = 16 + 24 = 40 and \displaystyle \sqrt{40} = 2\sqrt{10}, so \displaystyle x = \frac{-4 \pm 2\sqrt{10}}{4}; cancel the common 2.
\displaystyle n = \frac{1 \pm \sqrt{6}}{5}. b^2 - 4ac = 4 + 20 = 24 and \displaystyle \sqrt{24} = 2\sqrt{6}, so \displaystyle n = \frac{2 \pm 2\sqrt{6}}{10}; cancel the common 2.
No real solutions. b^2 - 4ac = 1 - 12 = -11, which is negative. Over the complex numbers, \displaystyle x = \frac{1 \pm i\sqrt{11}}{2}.
x = 0 or x = 8. No constant term, so factor: x(x - 8) = 0.
\displaystyle x = -2 \pm 3\sqrt{2}. Square roots: \displaystyle x + 2 = \pm\sqrt{18} = \pm 3\sqrt{2}.
\displaystyle x = \frac{5 \pm \sqrt{37}}{6}. The discriminant is 25 + 12 = 37, not a perfect square, so use the formula.
\displaystyle x = -3 \pm \sqrt{5}. Leading coefficient 1, even b: complete the square, x^2 + 6x + 9 = 5, so (x + 3)^2 = 5.
\displaystyle x = -\frac{1}{2} or x = -3. The discriminant is 49 - 24 = 25 = 5^2, so factor: (2x + 1)(x + 3) = 0.
x = 4 or x = -2. The only x is in the bracket, so square roots, once the square is alone: add 7, then divide by 3, to get (x - 1)^2 = 9, so x - 1 = \pm 3.
Factors; \displaystyle x = \frac{5}{2} or \displaystyle x = -\frac{4}{3}. The discriminant is 49 + 480 = 529 = 23^2, and (2x - 5)(3x + 4) = 0.
Does not factor; \displaystyle x = \frac{-5 \pm \sqrt{73}}{8}. The discriminant is 25 + 48 = 73.
Factors; \displaystyle x = \frac{2}{3}, the only solution. The discriminant is 144 - 144 = 0, and the trinomial is the perfect square (3x - 2)^2.
\displaystyle x = -4 \pm 3\sqrt{2}. Standard form x^2 + 8x - 2 = 0; b^2 - 4ac = 64 + 8 = 72 and \displaystyle \sqrt{72} = 6\sqrt{2}, so \displaystyle x = \frac{-8 \pm 6\sqrt{2}}{2}.
x = 1 or x = -1. Expand: 2x^2 + 3x - 2 = 3x, so 2x^2 - 2 = 0 with b = 0; b^2 - 4ac = 0 + 16 = 16, so \displaystyle x = \frac{0 \pm 4}{4}. Square roots get there faster: x^2 = 1.
x = 3 or x = -1. The square root property does not apply, since x also appears on the right. Expand: 4x^2 - 4x + 1 = 2x^2 + 7, so 2x^2 - 4x - 6 = 0, and dividing by 2 gives (x - 3)(x + 1) = 0.
\displaystyle x = \frac{-1 \pm \sqrt{19}}{3}. Multiply by 6: 3x^2 + 2x - 6 = 0; b^2 - 4ac = 4 + 72 = 76 and \displaystyle \sqrt{76} = 2\sqrt{19}, so \displaystyle x = \frac{-2 \pm 2\sqrt{19}}{6}.
\displaystyle x = \frac{1}{5} or x = -1. Multiply by 10: 5x^2 + 4x - 1 = 0; b^2 - 4ac = 16 + 20 = 36, a perfect square, so it factors as (5x - 1)(x + 1) = 0, and the formula agrees: \displaystyle x = \frac{-4 \pm 6}{10}.
The width is \displaystyle \sqrt{11} - 1 m, about 2.32 m. With width w, w(w + 2) = 10, so w^2 + 2w - 10 = 0; b^2 - 4ac = 4 + 40 = 44 and \displaystyle \sqrt{44} = 2\sqrt{11}, so \displaystyle w = -1 \pm \sqrt{11}. The negative solution is not a width.
At t = 1 and t = 3 seconds, once going up and once coming down. -5t^2 + 20t + 1 = 16 gives -5t^2 + 20t - 15 = 0; divide by -5: t^2 - 4t + 3 = 0, and \displaystyle t = \frac{4 \pm \sqrt{4}}{2} = \frac{4 \pm 2}{2}.
After t = 3 seconds. No t term, so square roots: 16t^2 = 144, t^2 = 9, t = \pm 3, and a time after the drop is positive.
After \displaystyle t = \frac{4 + \sqrt{21}}{4} \approx 2.15 seconds. The discriminant of -16t^2 + 32t + 5 = 0 is 1024 + 320 = 1344 = 64 \cdot 21, not a perfect square, so the formula gives \displaystyle t = \frac{-32 \pm 8\sqrt{21}}{-32} = 1 \mp \frac{\sqrt{21}}{4}. The negative solution, about -0.15, is before the throw.
At \displaystyle x = 2 \pm \sqrt{3}, about 3.73 and 0.27. The parabola meets the x-axis where y = 0, so solve x^2 - 4x + 1 = 0: b^2 - 4ac = 16 - 4 = 12 and \displaystyle \sqrt{12} = 2\sqrt{3}, so \displaystyle x = \frac{4 \pm 2\sqrt{3}}{2}. The two crossings sit \displaystyle \sqrt{3} either side of the axis of symmetry x = 2.
The formula solves ax^2 + bx + c = 0, with zero on the right. In standard form the equation is x^2 + 3x - 4 = 0, so c = -4, and its solutions are 1 and -4. Using c = 4 solves x^2 + 3x + 4 = 0 instead, a different equation, which has no real solutions at all.
The two numerators are \displaystyle -b + \sqrt{b^2 - 4ac} and \displaystyle -b - \sqrt{b^2 - 4ac}, over the same denominator 2a. Adding them, the square roots cancel and leave -2b, so the sum is \displaystyle \frac{-2b}{2a} = -\frac{b}{a}.
If a and c have opposite signs then ac < 0, so -4ac > 0. Since b^2 \ge 0 too, b^2 - 4ac > 0. A positive number has a positive square root, and adding it and subtracting it give two different real values.
The fraction bar groups the whole numerator, so the 6 divides both terms: \displaystyle \frac{-6}{6} \pm \frac{2\sqrt{15}}{6} = -1 \pm \frac{\sqrt{15}}{3}. Getting \displaystyle -1 \pm 2\sqrt{15} would mean dividing the -6 by 6 and leaving \displaystyle 2\sqrt{15} undivided, which divides only part of the numerator and changes its value.
The solutions are the numbers that make the equation true, a property of the equation alone. Every step of either method replaces the equation by one with exactly the same solutions, so both chains end at the same set of numbers. The formula is no exception, since it is completing the square done once on the letters a, b and c.
x = 0 or x = 5, from x^2 - 5x = 0, that is x(x - 5) = 0. Dividing by x is allowed only when x \neq 0, so it assumes away the solution x = 0 and leaves only x = 5.
With c = 0 the discriminant is b^2 - 4a \cdot 0 = b^2, the square of the integer b. By the factoring test the trinomial factors, and indeed ax^2 + bx = x(ax + b).
Completing the square adds \displaystyle \left(\frac{b}{2}\right)^2, and when b is even, \displaystyle \frac{b}{2} is a whole number, so no fraction appears. Moving c across and adding \displaystyle \left(\frac{b}{2}\right)^2 to both sides gives \displaystyle \left(x + \frac{b}{2}\right)^2 = \frac{b^2}{4} - c; taking square roots with a \pm and subtracting \displaystyle \frac{b}{2} gives the formula stated.