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Elan Coil [88]
2 years ago
8

How do you solve this ? Check for extraneous solutions

Mathematics
1 answer:
kakasveta [241]2 years ago
3 0

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.

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