Quadratic Formula and Discriminant, SAT math lesson

Factoring is fast when it works, but an equation like $x^2 - 4x - 1 = 0$ has no nice factors at all. Completing the square always works, and if you do it once with letters instead of numbers you get a formula that solves every quadratic equation. Even better, one small piece of that formula answers a question the SAT loves to ask without any solving, namely how many real solutions there are.

See also The Quadratic Formula and Real Solutions in Foundations for Calculus.

Key terms

Leading coefficient

Definition A quadratic equation in standard form is

ax^2 + bx + c = 0,

where $a$ , $b$ and $c$ are real constants and $a \ne 0$ . The number $a$ , the coefficient of the highest power of $x$ , is the leading coefficient. If $a = 0$ , the equation is linear, not quadratic, and nothing in this lesson applies to it.

In plain words Before using any quadratic rule, move everything to one side, order the terms by power, and read off $a$ , $b$ and $c$ with their signs.

Example The equation $5 + 2x = 3x^2$ becomes $-3x^2 + 2x + 5 = 0$ after moving every term to the left side. So $a = -3$ , $b = 2$ and $c = 5$ . Multiplying through by $-1$ gives the equally valid $3x^2 - 2x - 5 = 0$ , with $a = 3$ , $b = -2$ and $c = -5$ , and the same solutions.

Quadratic formula

Definition If $a \ne 0$ , the solutions of $ax^2 + bx + c = 0$ are

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

These are real numbers when $b^2 - 4ac \ge 0$ . When $b^2 - 4ac < 0$ the square root is not a real number, and the equation has no real solutions. The formula comes from completing the square in the general equation. The sign $\pm$ stands for the two branches, one with $+$ and one with $-$ . Learn the formula by heart rather than counting on looking it up during the test.

In plain words Plug in $a$ , $b$ and $c$ , and the formula hands you both solutions at once. Put each substituted number in parentheses, especially negative ones, before doing any arithmetic.

Example Solve $2x^2 - 7x + 3 = 0$ . Here $a = 2$ , $b = -7$ and $c = 3$ .

\begin{aligned} x &= \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(3)}}{2(2)} \\[0.4em] &= \frac{7 \pm \sqrt{25}}{4} = \frac{7 \pm 5}{4}. \end{aligned}

The plus branch gives $x = 3$ and the minus branch gives $x = \frac{1}{2}$ . Both roots are rational because the coefficients are integers and $25$ is a perfect square, and indeed the left side factors as $(2x - 1)(x - 3)$ .

Discriminant

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

D = b^2 - 4ac.

It decides how many real solutions the equation has.

Discriminant	Real solutions	Graph of $y = ax^2 + bx + c$
$D > 0$	two distinct	crosses the $x$ -axis twice
$D = 0$	one repeated	touches the $x$ -axis once
$D < 0$	none	misses the $x$ -axis

When $D < 0$ there are no real solutions, but there are still two complex solutions, so do not say the equation has no solutions of any kind. On the SAT variables are real unless a question says otherwise, so “no real solutions” is the right reading there.

In plain words The discriminant is the part under the square root. A positive number has two square roots, zero has one, and a negative number has no real square root.

Example For $3x^2 - 5x + 4 = 0$ ,

\begin{aligned} D &= (-5)^2 - 4(3)(4) \\[0.4em] &= 25 - 48 = -23. \end{aligned}

Since $D < 0$ , the equation has no real solutions, and the parabola $y = 3x^2 - 5x + 4$ stays entirely above the $x$ -axis. You found that without solving anything.

Real root

Definition A real root (or real solution) of $ax^2 + bx + c = 0$ is a real number $r$ with $ar^2 + br + c = 0$ . Each real root $r$ is the $x$ -coordinate of an $x$ -intercept $(r, 0)$ of the graph of $y = ax^2 + bx + c$ . When $D > 0$ there are exactly two distinct real roots. If $a$ , $b$ and $c$ are integers and $D \ge 0$ , the roots are rational exactly when $D$ is a perfect square such as $0$ , $1$ , $4$ or $25$ . Otherwise both roots are irrational.

In plain words A real root is an ordinary number that makes the equation true. On a graph it is where the parabola meets the $x$ -axis.

Example Solve $x^2 - 4x - 1 = 0$ . Here $D = 16 + 4 = 20$ , which is positive but not a perfect square.

\begin{aligned} x &= \frac{4 \pm \sqrt{20}}{2} = \frac{4 \pm 2\sqrt{5}}{2} \\[0.4em] &= 2 \pm \sqrt{5}. \end{aligned}

The two real roots are about $4.236$ and $-0.236$ . In the Desmos graphing calculator you can type $y = x^2 - 4x - 1$ and click the curve. Gray points appear at features such as the $x$ -intercepts, and clicking one shows its coordinates as decimals. On a multiple choice question, though, the answer is usually listed in the exact form $2 \pm \sqrt{5}$ .

Repeated root

Definition A quadratic equation has a repeated root (a double root) $r$ when it can be written as $a(x - r)^2 = 0$ with $a \ne 0$ . This happens exactly when $D = 0$ , and then the only solution is

x = -\frac{b}{2a}.

The equation has one real solution, not two, even though the factor $x - r$ appears twice.

In plain words With a zero discriminant the $\pm$ adds and subtracts nothing, so both branches land on the same number. The parabola just touches the $x$ -axis at its vertex.

Example For $4x^2 - 12x + 9 = 0$ the discriminant is $144 - 4(4)(9) = 0$ . The repeated root is

x = \frac{12}{2 \cdot 4} = \frac{3}{2}.

The left side is $(2x - 3)^2$ , a perfect square, which confirms that $\frac{3}{2}$ is the one and only solution.

Check yourself

How many distinct real solutions does the equation $2x^2 + 5x + 4 = 0$ have? (A) Zero, (B) Exactly one, (C) Exactly two, (D) Infinitely many

Show the answer

Choice (A) is correct. The discriminant is

D = 5^2 - 4(2)(4) = 25 - 32 = -7,

which is negative, so there are no real solutions. Choice (C) is what you get by computing $25 + 32$ , which drops the minus sign in $b^2 - 4ac$ . Choice (B) mistakes a negative discriminant for a repeated root, but exactly one solution needs $D = 0$ , and a quadratic equation never has infinitely many solutions, which rules out (D).

The solutions of $x^2 - 6x + 4 = 0$ can be written as $x = 3 \pm \sqrt{k}$ , where $k$ is a constant. What is the value of $k$ ?

Show the answer

The answer is $5$ . The quadratic formula gives

x = \frac{6 \pm \sqrt{36 - 16}}{2} = \frac{6 \pm \sqrt{20}}{2}.

Since $\sqrt{20} = 2\sqrt{5}$ , this is $3 \pm \sqrt{5}$ . Entering $20$ is the trap, because the $2$ in the denominator still has to divide the square root.

Which equation has exactly one real solution? (A) $x^2 - 25 = 0$ , (B) $x^2 + 25 = 0$ , (C) $x^2 - 10x + 25 = 0$ , (D) $x^2 - 10x - 25 = 0$

Show the answer

Choice (C) is correct. Its discriminant is $100 - 100 = 0$ , and it equals $(x - 5)^2 = 0$ with the repeated root $5$ . Choice (A) has $D = 100$ and two solutions, $5$ and $-5$ . Choice (B) has $D = -100$ and no real solutions. Choice (D) has $D = 100 + 100 = 200$ and two irrational solutions.

Quadratic Formula and Discriminant

Theory, part 51

Practice, part 52

Key terms

Leading coefficient

Quadratic formula

Discriminant

Real root

Repeated root

Check yourself

How this lesson connects

Builds on

Used later in