Question

5.3 - online quiz - solving quadratic equations by fa
started: dec 12 at 6:18pm
quiz instructions
question 2
solve the following equation by factoring.
$x^2 + 7x = 8$
\\(\circ\\) x = -1, x = 8
\\(\circ\\) x = -1, x = -8
\\(\circ\\) x = 1, x = 8
\\(\circ\\) x = 1, x = -8

Explanation:

Step1: Rearrange the equation

First, we need to rewrite the quadratic equation in standard form \(ax^{2}+bx + c = 0\). Subtract 8 from both sides of the equation \(x^{2}+7x = 8\), we get \(x^{2}+7x - 8=0\).

Step2: Factor the quadratic expression

We need to find two numbers that multiply to -8 (the product of \(a = 1\) and \(c=-8\)) and add up to 7 (the coefficient of \(x\)). The numbers are 8 and - 1, because \(8\times(-1)=-8\) and \(8+(-1) = 7\). So we can factor the quadratic as \((x + 8)(x - 1)=0\).

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:

• For \(x + 8=0\), subtract 8 from both sides, we get \(x=-8\).
• For \(x - 1=0\), add 1 to both sides, we get \(x = 1\).

Answer:

\(x = 1,x=-8\) (the option: \(x = 1,x=-8\))