The Quadratic Formula and Real Solutions, foundations for calculus lesson

Throw a ball upward and ask when it lands, or set a derivative equal to zero to find where a graph turns. Both questions often end in an equation like $ax^2 + bx + c = 0$ . The quadratic formula solves every such equation, and before you do any arithmetic, one number inside it already tells you how many real answers to expect.

Key terms

Quadratic equation

Definition A quadratic equation in $x$ is an equation that can be written in the standard form

ax^2 + bx + c = 0,

where $a$ , $b$ and $c$ are real numbers and $a \ne 0$ . If $a = 0$ the equation is $bx + c = 0$ , which is linear, and the quadratic formula does not apply to it.

In plain words A quadratic equation has an $x^2$ term and no higher power. Before solving, move everything to one side so the other side is $0$ .

Example The equation $3x^2 = 5x + 2$ becomes $3x^2 - 5x - 2 = 0$ after subtracting $5x + 2$ from both sides. Its solutions turn out to be $2$ and $-\frac{1}{3}$ .

Coefficients

Definition In the standard form $ax^2 + bx + c = 0$ , the number $a$ is the coefficient of $x^2$ , $b$ is the coefficient of $x$ , and $c$ is the constant term. Each coefficient includes its sign, and a missing term has coefficient $0$ .

In plain words Read the three numbers from the standard form, signs and all. The order of the terms on the page does not matter, only which power of $x$ each number goes with.

Example The equation $2 - x^2 + 4x = 0$ has $a = -1$ , $b = 4$ and $c = 2$ . The equation $5x^2 - 7 = 0$ has $a = 5$ , $b = 0$ and $c = -7$ .

Quadratic formula

Definition Let $a$ , $b$ , $c$ be real with $a \ne 0$ . If $b^2 - 4ac \ge 0$ , the real solutions of $ax^2 + bx + c = 0$ are exactly

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

The formula comes from completing the square. Dividing by $a$ and completing the square gives

\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2},

and taking square roots of both sides, which is possible in the real numbers exactly when the right side is nonnegative, gives the formula. If $b^2 - 4ac < 0$ there is no real solution. See completing the square for that step.

In plain words Put the three coefficients into the formula and simplify. The whole top line, including $-b$ , is divided by $2a$ .

Example Solve $2x^2 - 3x - 5 = 0$ , where $a = 2$ , $b = -3$ and $c = -5$ . Then $b^2 - 4ac = 9 + 40 = 49$ , so

x = \frac{3 \pm 7}{4}, \qquad x = \frac{5}{2} \ \text{or} \ x = -1.

Check $x = -1$ in the original, $2 + 3 - 5 = 0$ .

Plus or minus

Definition The symbol $\pm$ in the formula stands for two separate expressions, one with $+$ and one with $-$ ,

\begin{gathered} x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a}, \\[0.4em] x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a}. \end{gathered}

It comes from the fact that, for $d > 0$ , the equation $u^2 = d$ has the two solutions $u = \sqrt{d}$ and $u = -\sqrt{d}$ . In each case the complete numerator is divided by $2a$ . Writing $-b \pm \frac{\sqrt{b^2 - 4ac}}{2a}$ is in general a different number and is wrong.

In plain words One formula, two answers. Do the calculation once with plus and once with minus.

Example For $x^2 - 4x + 1 = 0$ we have $b^2 - 4ac = 16 - 4 = 12$ and $\sqrt{12} = 2\sqrt{3}$ , so

x = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}.

Both terms of the numerator were divided by $2$ . The solutions are about $3.732$ and $0.268$ .

Discriminant

Definition The discriminant of $ax^2 + bx + c$ with $a \ne 0$ is the number

D = b^2 - 4ac.

If $D > 0$ the equation has two distinct real solutions. If $D = 0$ it has exactly one real solution, $x = -\frac{b}{2a}$ . If $D < 0$ it has no real solution.

In plain words The discriminant is the number under the square root. Its sign decides how many real answers there are, before you compute any of them.

Example The three equations below differ only in the constant term.

Equation	$D$	Real solutions
$x^2 + 2x - 3 = 0$	$16$	two, $x = 1$ and $x = -3$
$x^2 + 2x + 1 = 0$	$0$	one, $x = -1$
$x^2 + 2x + 5 = 0$	$-16$	none

On the graphs, the parabola $y = x^2 + 2x + c$ crosses the $x$ axis twice, touches it once, or stays above it.

Real solution

Definition A real solution of an equation is a real number that makes the equation true. When $D < 0$ , the right side of $\left(x + \frac{b}{2a}\right)^2 = \frac{D}{4a^2}$ is negative while the left side is a real square and so is nonnegative. No real $x$ can satisfy it, so the quadratic has no real solution. This course works over the real numbers, so in that case the answer is simply that there is no solution. The complex numbers, which this course does not use, are where such equations do have solutions.

In plain words A negative discriminant means you would need the square root of a negative number. No real number squares to a negative, so there is nothing to find on the number line.

Example For $x^2 + x + 1 = 0$ the discriminant is $1 - 4 = -3 < 0$ . Completing the square confirms it,

x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4} \ge \frac{3}{4},

so the left side is never $0$ for real $x$ .

Repeated root

Definition When $D = 0$ , the quadratic factors as

ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2,

so $x = -\frac{b}{2a}$ is its only root. It is called a repeated root, or a double root, because the factor $x + \frac{b}{2a}$ appears twice. There is one distinct real solution, not two.

In plain words The two answers from plus and minus collapse into the same number, since you add and subtract zero. The parabola just touches the axis at its vertex.

Example For $4x^2 - 12x + 9 = 0$ the discriminant is $144 - 144 = 0$ . The formula gives $x = \frac{12}{8} = \frac{3}{2}$ , and indeed $4x^2 - 12x + 9 = (2x - 3)^2$ .

Check yourself

Solve $3x^2 + 2x - 4 = 0$ exactly.

Show the answer

Here $a = 3$ , $b = 2$ , $c = -4$ , so $D = 4 + 48 = 52$ and $\sqrt{52} = 2\sqrt{13}$ . Then

x = \frac{-2 \pm 2\sqrt{13}}{6} = \frac{-1 \pm \sqrt{13}}{3}.

These are about $0.869$ and $-1.535$ .

Explain without solving why $2x^2 - 4x + 5 = 0$ has no real solution.

Show the answer

The discriminant is $(-4)^2 - 4 \cdot 2 \cdot 5 = 16 - 40 = -24$ , which is negative. A real solution would require a real square root of $-24$ , and none exists.

For which real numbers $k$ does $x^2 + kx + 9 = 0$ have a repeated root?

Show the answer

A repeated root needs $D = k^2 - 36 = 0$ , so $k = 6$ or $k = -6$ . For $k = 6$ the root is $-3$ , since $x^2 + 6x + 9 = (x + 3)^2$ , and for $k = -6$ it is $3$ .

The Quadratic Formula and Real Solutions

Key terms

Quadratic equation

Coefficients

Quadratic formula

Plus or minus

Discriminant

Real solution

Repeated root

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus