Before you read

Answer first — see what you already know.

A bridge arch follows $x^{2} - 4 x + 3 = 0$ . Use the quadratic formula to find where it meets the roadway.

A
$x = 1$ or $x = 3$
B
$x = - 1$ or $x = - 3$
C
$x = 2$ or $x = 6$
D
$x = 4$ or $x = 3$

Check your answer

Answer:

Before you read

Answer first — see what you already know.

An engineering tolerance gives $x^{2} + 2 x - 5 = 0$ . What are the solutions in simplified form?

A
$x = \frac{- 2 \pm 24}{2}$
B
$x = - 1 \pm 6$
C
$x = \frac{2 \pm 24}{2}$
D
$x = 1 \pm 6$

Check your answer

Answer:

Theory

The Formula

For any quadratic equation $ax^2 + bx + c = 0$ (where $a \neq 0$ ):

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

$ax^2 + bx + c = 0$
$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ quadratic formula

The quadratic formula

This formula always works, whether the quadratic factors or not.

Memorize it. It appears on nearly every SAT.

The $\pm$ means there are (usually) two solutions:
$x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \qquad x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a}$

x² + 2x - 6: solutions from the quadratic formula

Theory

Using the Formula Step by Step

1. Write the equation in standard form: $ax^2 + bx + c = 0$
2. Identify $a$ , $b$ , and $c$
3. Compute the discriminant: $\Delta = b^2 - 4ac$
4. Substitute into the formula
5. Simplify

Example 1

Solve $x^2 + 6x + 5 = 0$

$a = 1$ , $b = 6$ , $c = 5$

$\Delta = 6^2 - 4(1)(5) = 36 - 20 = 16$

$x = \frac{-6 \pm \sqrt{16}}{2(1)} = \frac{-6 \pm 4}{2}$

$x_1 = \frac{-6+4}{2} = \frac{-2}{2} = -1$

$x_2 = \frac{-6-4}{2} = \frac{-10}{2} = -5$

x = -1

x = -5

Example 2

Solve $2x^2 - 3x - 1 = 0$

$a = 2$ , $b = -3$ , $c = -1$

$\Delta = (-3)^2 - 4(2)(-1) = 9 + 8 = 17$

$x = \frac{-(-3) \pm \sqrt{17}}{2(2)} = \frac{3 \pm \sqrt{17}}{4}$

This doesn't simplify further.

x = \frac{3 \pm \sqrt{17}}{4}

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