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Chapter 0: Problem 67

Solve each quadratic equation by completing the square. $$x^{2}+6 x=7$$

Short Answer

Expert verified
The solutions to the quadratic equation are \(x=1\) and \(x=-7\).

Step by step solution

01

Rearrange the equation

Rewrite the equation as \(x^{2}+6x-7=0\). This is done to bring the equation to the standard quadratic form \(ax^{2}+bx+c=0\), where a, b, and c are coefficients.
02

Complete the square

Complete the square on the left side of the equation and balance this by adding the same value to the right side. Use the formula \(x^{2}+bx+c={((b/2)^{2})}\) where \(c={(b/2)^{2}}\), and \(b\) is the coefficient of \(x\). Here, \(b=6\), so \(c={(3)^{2}}=9\). So, the equation becomes \(x^{2}+6x+9=7+9\) or \(x^{2}+6x+9=16\).
03

Factorise the equation

Factor the left side of the equation to form a perfect square. We obtain \((x+3)^{2}=16\).
04

Find the square root

Calculate the square root of both sides of the equation. This gives \(x+3=\pm4\).
05

Isolate x

Solve for \(x\) by subtracting 3 from both sides. This gives \(x=4-3=1\) or \(x=-4-3=-7\).

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Most popular questions from this chapter

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. $$6+\frac{1}{x}=\frac{7}{x}$$

What does it mean to solve a formula for a variable?

Describe two ways to simplify \(\frac{\frac{3}{x}+\frac{2}{x^{2}}}{\frac{1}{x^{2}}+\frac{2}{x}}\).

Describe ways in which solving a linear inequality is different than solving a linear equation.

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. $$\frac{x^{2}-25}{x-5}=x-5$$
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