Question

what are the solutions of $x^2 + 6x - 6 = 10$?
$\bigcirc$ $x = -11$ or $x = 1$
$\bigcirc$ $x = -11$ or $x = -1$
$\bigcirc$ $x = -8$ or $x = -2$
$\bigcirc$ $x = -8$ or $x = 2$

Explanation:

Step1: Rearrange the equation

First, we need to set the equation \(x^{2}+6x - 6=10\) to standard quadratic form \(ax^{2}+bx + c = 0\). Subtract 10 from both sides:
\(x^{2}+6x-6 - 10=0\)
Simplify to get:
\(x^{2}+6x - 16=0\)

Step2: Factor the quadratic equation

We need to find two numbers that multiply to \(- 16\) and add up to \(6\). The numbers are \(8\) and \(-2\) because \(8\times(-2)=-16\) and \(8+( - 2)=6\). So we can factor the quadratic as:
\((x + 8)(x - 2)=0\)

Step3: Solve for x

Using the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\). So we set each factor equal to zero:
Case 1: \(x+8 = 0\), then \(x=-8\)
Case 2: \(x - 2=0\), then \(x = 2\)

Answer:

\(x=-8\) or \(x = 2\) (corresponding to the option "x = - 8 or x = 2")