Question

name: ______________________ unit 4: solving quadratic equations
date: __________________ per: __ homework 8: the quadratic formula
this is a 2 - page document!
directions: solve each equation by the quadratic formula.

1. $x^2 + 12x - 8 = 0$

$x = \\{ \square \\}$

1. $-2x^2 + 7x = 3$

$x = \\{ \square, \square \\}$

Explanation:

Step1: Recall Quadratic Formula

The quadratic formula for a quadratic equation \(ax^2 + bx + c = 0\) is \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\).

Step2: Solve Equation 1: \(x^{2}+12x - 8 = 0\)

For \(x^{2}+12x - 8 = 0\), we have \(a = 1\), \(b = 12\), \(c=-8\).
Substitute into the formula:
\(x=\frac{-12\pm\sqrt{12^{2}-4(1)(-8)}}{2(1)}=\frac{-12\pm\sqrt{144 + 32}}{2}=\frac{-12\pm\sqrt{176}}{2}=\frac{-12\pm4\sqrt{11}}{2}=-6\pm2\sqrt{11}\)

Step3: Solve Equation 2: \(-2x^{2}+7x = 3\)

First, rewrite in standard form: \(-2x^{2}+7x - 3 = 0\) (or multiply by -1: \(2x^{2}-7x + 3 = 0\)). Let's use \(2x^{2}-7x + 3 = 0\), so \(a = 2\), \(b=-7\), \(c = 3\).
Substitute into the formula:
\(x=\frac{7\pm\sqrt{(-7)^{2}-4(2)(3)}}{2(2)}=\frac{7\pm\sqrt{49 - 24}}{4}=\frac{7\pm\sqrt{25}}{4}=\frac{7\pm5}{4}\)
This gives two solutions:
\(x_1=\frac{7 + 5}{4}=\frac{12}{4}=3\) and \(x_2=\frac{7 - 5}{4}=\frac{2}{4}=\frac{1}{2}\)

Answer:

1. \(x=\{-6 + 2\sqrt{11},-6 - 2\sqrt{11}\}\)
2. \(x=\{\frac{1}{2},3\}\)