ACT Math · Algebra · free lesson

Quadratic Equations

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The idea

What a quadratic equation is

A quadratic equation has the variable squared and no higher power. Its standard form is $ax^{2} + bx + c = 0$ with $a \ne 0$ . It usually has two solutions, also called roots. This lesson gives four ways to find them: factoring, square roots, completing the square and the quadratic formula. It also shows how to tell how many roots there are without finding them.

The zero-product rule

Rule.

If a product of two factors is 0, then at least one of the factors is 0. So $(x - 3)(x + 5) = 0$ means $x - 3 = 0$ or $x + 5 = 0$ : $x = 3$ or $x = -5$ .

This works only when the other side is 0, so the first step is always to move every term to one side. Notice the signs: the factor $x + 5$ gives the root $-5$ .

Solving by factoring

To factor $x^2 - 2x - 15$ , look for two numbers whose product is $-15$ and whose sum is $-2$ . They are $-5$ and 3. The engine factors the expression and solves the equation:

\begin{aligned} y &= x^{2} - 2x - 15 \\ &= (x + 3)(x - 5) \\ x^{2} - 2x - 15 &= 0 \\ x &= [5, -3] \end{aligned}

The roots are where the graph of $y = x^2 - 2x - 15$ , a parabola, crosses the $x$ -axis: there $y$ is 0. Watch the curve draw itself from left to right. It comes down, crosses the axis at the first root, turns, and crosses again at the second.

When $a$ is not 1, the factors have numbers in front of $x$ . For $2x^2 + x - 6 = 0$ the engine finds the factors; multiplying them out gives back the left side, and each factor gives one root:

\begin{aligned} F &= 2x^{2} + x - 6 \\ &= (x + 2)(2x - 3) \\ p &= F \\ &= (x + 2)(2x - 3) \\ &= 2x^{2} + x - 6 \\ 2x^{2} + x - 6 &= 0 \\ x &= [\frac{3}{2}, -2] \end{aligned}

Three special products are worth knowing on sight:

Pattern	Factored	Example
$a^{2} - b^{2}$	$(a - b)(a + b)$	$x^2 - 49 = (x - 7)(x + 7)$
$a^2 + 2ab + b^2$	$(a + b)^{2}$	$x^2 + 10x + 25 = (x + 5)^2$
$a^2 - 2ab + b^2$	$(a - b)^{2}$	$4x^2 - 12x + 9 = (2x - 3)^2$

Solving by square roots

When the equation is a square equal to a number, get the square alone and take the square root of both sides. Write both signs: two numbers have each positive square. For $2(x - 3)^2 - 7 = 43$ :

\begin{aligned} s &= \frac{43 + 7}{2} \\ &= 25 \\ (x - 3)^{2} &= s \\ (x - 3)^{2} &= 25 \\ x &= [8, -2] \end{aligned}

$(x - 3)^{2} = 25$ , so $x - 3$ is 5 or $-5$ , and $x$ is $8$ or $-2$ .

Warning.

Never divide both sides by $x$ . In $x^2 = 5x$ , dividing by $x$ leaves only $x = 5$ and loses the root 0. Write $x^2 - 5x = 0$ , so $x(x - 5) = 0$ , and the roots are 0 and 5.

Completing the square

Any quadratic can be turned into a square equal to a number. For $x^2 + 6x - 7 = 0$ , move the 7 across and add the square of half of 6 to both sides:

\begin{aligned} R &= 7 + (\frac{6}{2})^{2} \\ &= 16 \\ (x + 3)^{2} &= R \\ (x + 3)^{2} &= 16 \\ x &= [1, -7] \end{aligned}

The picture shows where the name comes from. $x^2 + 6x$ is a square of side $x$ and two strips, each 3 wide. Together they make a larger square with one corner missing; the 3 by 3 corner, 9, completes the square of side $x + 3$ . Watch it build piece by piece.

The equation becomes $(x + 3)^{2} = 16$ , so $x + 3 = \pm 4$ . Doing the same to $ax^{2} + bx + c = 0$ with letters gives a formula that works for every quadratic.

The quadratic formula

Rule.

The solutions of $ax^{2} + bx + c = 0$ are

\begin{aligned} x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \end{aligned}

For $x^2 - 6x + 2 = 0$ , which does not factor with whole numbers, $a = 1$ , $b = -6$ and $c = 2$ . Work out the number under the root first:

\begin{aligned} D &= b^{2} - 4ac \\ &= (-6)^{2} - 4 \times 1 \times 2 \\ &= 28 \\ R &= \sqrt{D} \\ &= \sqrt{28} \\ &= 2\sqrt{7} \\ x^{2} - 6x + 2 &= 0 \\ x &= [3 + \sqrt{7}, 3 - \sqrt{7}] \end{aligned}

So $x = \frac{6 \pm 2\sqrt{7}}{2}$ , which is $3 \pm \sqrt{7}$ : about $5.646$ and $0.354$ . Divide both terms of the top by $2a$ ; dividing only one is a slip the ACT puts among the options.

The discriminant: two, one or no real roots

The number under the root, $b^2 - 4ac$ , is the discriminant. A positive number has two square roots, zero has one, and a negative number has no real square root. Drag the slider to change $c$ in $x^2 - 4x + c = 0$ . As $c$ grows, the parabola rises: it crosses the axis twice, then touches it at $c = 4$ , then lifts off it.

Rule.

For $ax^{2} + bx + c = 0$ : if $b^2 - 4ac > 0$ there are two distinct real roots; if $b^2 - 4ac = 0$ there is exactly one (a double root); if $b^2 - 4ac < 0$ there is no real root.

With no real root the formula still gives two answers: the square root of a negative number is imaginary (topic 05). For $x^2 - 4x + 5 = 0$ the discriminant and the two complex roots are

\begin{aligned} D &= (-4)^{2} - 4 \times 1 \times 5 \\ &= -4 \\ x^{2} - 4x + 5 &= 0 \\ z &= [2 + i, 2 - i] \end{aligned}

The roots come as a pair of conjugates, $a + bi$ and $a - bi$ . The ACT may ask for them in this form.

Sum and product of the roots

If the roots are $r$ and $s$ , then $a(x - r)(x - s)$ multiplied out is $ax^2 - a(r + s)x + ars$ . Matching it with $ax^{2} + bx + c$ :

Rule.

The sum of the roots is $-\frac{b}{a}$ and their product is $\frac{c}{a}$ .

For $2x^2 - 7x + 3 = 0$ the roots $r$ and $s$ , their sum $S$ and their product $P$ are:

\begin{aligned} 2x^{2} - 7x + 3 &= 0 \\ x &= [3, \frac{1}{2}] \\ r &= \operatorname{max}(x) \\ &= 3 \\ s &= \operatorname{min}(x) \\ &= \frac{1}{2} \\ S &= r + s \\ &= 3 + \frac{1}{2} \\ &= \frac{7}{2} \\ P &= rs \\ &= 3 \times \frac{1}{2} \\ &= \frac{3}{2} \end{aligned}

The rule gives the same two numbers straight from $a = 2$ , $b = -7$ and $c = 3$ , without solving:

\begin{aligned} S &= \frac{-(-7)}{2} \\ &= \frac{7}{2} \\ P &= \frac{3}{2} \end{aligned}

The rule also runs backwards: an equation with the roots 4 and $-1$ is $(x - 4)(x + 1) = 0$ , and multiplied out it is

\begin{aligned} y &= (x - 4)(x + 1) \\ &= x^{2} - 3x - 4 \end{aligned}

Its $x$ coefficient is minus the sum of the roots, and its constant is their product.

Note.

On the ACT a quadratic question asks you to factor a trinomial, to give the solutions (or the sum of the solutions), to solve for a letter that is squared, to say how many real solutions there are, or to set up a quadratic from an area or a height and solve it. Get 0 on one side first, then choose the method: factor when two whole numbers work, square roots when there is no $x$ term, the formula otherwise.

Worked examples

Each example is one of the kinds of question the ACT asks about quadratic equations. The numbers in the working are found by the engine.

Example.

1. Factor and solve, with a number in front. Solve $3x^2 - 10x - 8 = 0$ .

Look for factors $(3x + p)(x + q)$ whose product is the left side. The constant $pq$ must be $-8$ and the middle term $3q + p$ must be $-10$ . The engine finds the pair; multiply out to check, then let each factor give a root:

\begin{aligned} F &= 3x^{2} - 10x - 8 \\ &= (3x + 2)(x - 4) \\ y &= F \\ &= (3x + 2)(x - 4) \\ &= 3x^{2} - 10x - 8 \\ 3x^{2} - 10x - 8 &= 0 \\ x &= [4, -\frac{2}{3}] \\ r &= \operatorname{max}(x) \\ &= 4 \\ s &= \operatorname{min}(x) \\ &= -\frac{2}{3} \end{aligned}

The roots are $4$ and $-\frac{2}{3}$ . The factor $3x + 2$ gives $-\frac{2}{3}$ , not $-2$ : divide by the 3 as well. On the graph these are the two crossings of the $x$ -axis.

Example.

2. A difference of squares and a perfect square. Solve $4x^2 - 25 = 0$ and $x^2 - 14x + 49 = 0$ .

The first is $(2x)^2 - 5^2$ , a difference of squares; the second is $x^2 - 2(7)x + 7^2$ , a perfect square.

\begin{aligned} p &= 4x^{2} - 25 \\ &= (2x + 5)(2x - 5) \\ w &= p \\ &= (2x + 5)(2x - 5) \\ &= 4x^{2} - 25 \\ 4x^{2} - 25 &= 0 \\ u &= [\frac{5}{2}, -\frac{5}{2}] \\ q &= x^{2} - 14x + 49 \\ &= (x - 7)^{2} \\ x^{2} - 14x + 49 &= 0 \\ v &= 7 \end{aligned}

So $4x^2 - 25 = {(2x + 5)(2x - 5)}$ , as multiplying out checks, with the roots $\frac{5}{2}$ and $-\frac{5}{2}$ . The perfect square has one root, $7$ : both factors give the same number, and the parabola only touches the $x$ -axis there.

Example.

3. Solve for a squared letter. The kinetic energy of a moving object is $E = \frac{1}{2}mv^2$ . Solve for $v$ , with $v > 0$ , and find $v$ when $m = 4$ and $E = 50$ .

Undo the operations on $v$ in reverse order: multiply both sides by 2, divide by $m$ , then take the square root. Only the positive root is kept, since $v > 0$ .

\begin{aligned} v^{2} &= \frac{2E}{m}, \quad v = \sqrt{\frac{2E}{m}} \end{aligned}

With numbers, the engine solves the equation as given, and the formula agrees:

\begin{aligned} E &= \frac{1}{2}mv^{2} \\ 50 &= \frac{1}{2} \times 4v^{2} \\ v &= [-5, 5] \\ w &= \sqrt{\frac{2E}{m}} \\ &= \sqrt{\frac{2 \times 50}{4}} \\ &= 5 \end{aligned}

So $v = 5$ . Forgetting the 2, which gives $\sqrt{\frac{E}{m}}$ , and forgetting the square root, which gives $\frac{2E}{m}$ , are the usual wrong options.

Example.

4. The quadratic formula. Solve $2x^2 + 4x - 3 = 0$ .

It does not factor with whole numbers. Here $a = 2$ , $b = 4$ , $c = -3$ . The discriminant first, then its root:

\begin{aligned} D &= b^{2} - 4ac \\ &= 4^{2} - 4 \times 2 \times (-3) \\ &= 40 \\ R &= \sqrt{D} \\ &= \sqrt{40} \\ &= 2\sqrt{10} \\ 2x^{2} + 4x - 3 &= 0 \\ x &= [\frac{\sqrt{10} - 2}{2}, -\frac{2 + \sqrt{10}}{2}] \end{aligned}

So $x = \frac{-4 \pm {2\sqrt{10}}}{4}$ . Both terms of the top share a factor 2, and dividing them both by 4 gives the engine's roots, $\frac{-2 \pm \sqrt{10}}{2}$ . Since $c$ is negative, $-4ac$ is added: writing $16 - 24$ under the root is the slip to avoid.

Example.

5. A constant from the number of roots. For what value of $k > 0$ does $x^{2} + kx + 16 = 0$ have exactly one real solution? And for which values of $c$ does $3x^2 - 6x + c = 0$ have no real solution?

Exactly one solution means the discriminant is 0:

\begin{aligned} k^{2} - 4 \times 1 \times 16 &= 0 \\ k &= [8, -8] \end{aligned}

Of the two values, the positive one, $8$ , is asked for. Then $x^2 + {8}x + 16 = (x + 4)^2$ .

No real solution means the discriminant is negative. Find where it is 0, then test a value past it:

\begin{aligned} (-6)^{2} - 4 \times 3c &= 0 \\ c &= 3 \\ D &= (-6)^{2} - 4 \times 3(c + 1) \\ &= (-6)^{2} - 4 \times 3 \times (3 + 1) \\ &= -12 \end{aligned}

At $c = 3$ the discriminant is 0 (one root), and above it the discriminant is negative, so the answer is $c > 3$ . The picture shows the parabola touching the axis at $c = 3$ and lifting off for $c = 4$ .

Example.

6. Complex roots, and their sum and product. Find the roots of $x^2 + 6x + 13 = 0$ , and check their sum and product.

The discriminant decides the kind of root:

\begin{aligned} D &= 6^{2} - 4 \times 1 \times 13 \\ &= -16 \\ x^{2} + 6x + 13 &= 0 \\ z &= [-3 + 2i, -3 - 2i] \end{aligned}

The discriminant is negative, so there is no real root. The formula gives the same pair: the middle $-\frac{b}{2a}$ is the real part, and $\frac{\sqrt{-D}}{2a}$ is the imaginary part.

\begin{aligned} u &= \operatorname{complex}(-\frac{6}{2}, \frac{\sqrt{-D}}{2}) \\ &= \operatorname{complex}(-\frac{6}{2}, \frac{\sqrt{-(-16)}}{2}) \\ &= -3 + 2i \\ w &= \operatorname{conjugate}(u) \\ &= \operatorname{conjugate}(-3 + 2i) \\ &= -3 - 2i \end{aligned}

Their sum and product, worked with complex numbers (topic 05), are $-\frac{b}{a}$ and $\frac{c}{a}$ , as for real roots:

\begin{aligned} S &= \operatorname{add}(u, w) \\ &= \operatorname{add}(-3 + 2i, -3 - 2i) \\ &= -6 \\ P &= \operatorname{mul}(u, w) \\ &= \operatorname{mul}(-3 + 2i, -3 - 2i) \\ &= 13 \end{aligned}

Example.

7. A ball in the air. A ball is thrown upward from a roof. Its height in feet after $t$ seconds is $h = -16t^2 + 48t + 64$ . When does it hit the ground?

On the ground $h = 0$ . Divide the equation by $-16$ to get $t^2 - 3t - 4 = 0$ , then solve:

\begin{aligned} -16t^{2} + 48t + 64 &= 0 \\ t &= [-1, 4] \\ T &= \operatorname{max}(t) \\ &= 4 \end{aligned}

The two roots are $4$ and $-1$ , but time after the throw cannot be negative, so the ball lands after $4$ seconds. The graph draws the flight; the negative root lies off to the left, before the throw.

Practice

Answer ACT-style questions on quadratic equations. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise quadratic equations with new numbers