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The Quadratic Formula Solutions

Note

This deep dive into the quadratic formula is taken from “Extension 2 – The Quadratic Formula and the Discriminant” found in Algebra 1: Principles of Secondary Mathematics.

You already know how to solve a quadratic equation with these methods: 
- factoring  
- graphing  
- completing the square  
You can also solve quadratic equations for their roots using the quadratic formula . 
The quadratic formula for a quadratic equation in standard form is written as:
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}, where a \neq 0$
The quadratic formula is derived by taking the standard form of a quadratic equation and completing the square to solve for x. 
You can solve any quadratic equation in standard form using the quadratic formula. 
If using the quadratic formula to solve a quadratic equation results in a rational number , it means you also could have solved by factoring.

Example 1

Solve.

$4 x^{2} + 12 x + 9 = - 5$

Plan

Write in standard form

 Identify a, b, and c 

Substitute into formula

Implement

$4 x^{2} - 12 x + 14 = 0$

$a = 4, b = - 12, c = 14$

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

$x = \frac{- (- 12) \pm \sqrt{{(- 12)}^{2} - 4 (4) (14)}}{2 (4)}$

$x = \frac{12 \pm \sqrt{144 - 224}}{8} = \frac{12 \pm \sqrt{- 80}}{8} x = \frac{12 \pm i \sqrt{80}}{8} = \frac{12 \pm i \sqrt{16 \cdot 5}}{8} x = \frac{12 \pm 4 i \sqrt{5}}{8} = \frac{4 (3 \pm i \sqrt{5})}{4 (2)} x = \frac{3 \pm i \sqrt{5}}{2}$

Explain

Standard form
 Identify a, b, c 
Write the formula 
Substitute a, b, and c 
Simplify the right side

Example 2

Solve.

$5 x - 3 = 2 x^{2}$

$- 2 x^{2} + 5 x - 3 = 0 a = - 2, b = 5, c = - 3 x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} x = \frac{- (5) \pm \sqrt{{(5)}^{2} - 4 (- 2) (- 3)}}{2 (- 2)} = \frac{- 5 \pm \sqrt{25 - 24}}{- 4} x = \frac{- 5 \pm \sqrt{1}}{- 4} = \frac{- 5 \pm 1}{- 4} x = \frac{- 5 - 1}{- 4}, \frac{- 5 + 1}{- 4} x = \frac{3}{2}, 1$

Note

You could also rewrite the equation as $2 x^{2} - 5 x + 3 = 0$ and factor it, which would result in the same solution using $a = - 2, b = 5$ , and $c = - 3$ .