Question

How do you solve this ? Check for extraneous solutions

Answer 1

ANSWER

$x = 6$

EXPLANATION

The given equation is

$\sqrt{3x + 7} = x - 1$

We square both sides of the equation to obtain,

$(\sqrt{3x + 7} ) ^{2} =( x - 1)^{2}$

This implies that,

$3x + 7 = {x}^{2} - 2x + 1$

Rewrite in standard quadratic equation form.

${x}^{2} - 2x - 3x+ 1 - 7 = 0$

${x}^{2} -5x - 6= 0$

Factor

$(x - 6)(x + 1) = 0$

This implies that,

$x = 6 \: x = - 1$

We check for extraneous solution by substituting each x-value into the original equation.

When x=-1,

$\sqrt{3( - 1)+ 7} = - 1- 1$

$\sqrt{4} = - 2$

2=-2....False

Hence x=-1 is an extraneous solution.

When x=6,

$\sqrt{3( 6)+ 7} = 6- 1$

$\sqrt{25} = 5$

5=5 is True.

Hence x=6 is the only solution.