## given the equation Square root of x minus 5 + 7 = 11 Solve for x and identify if it is an extraneous solution.

Question

given the equation Square root of x minus 5 + 7 = 11 Solve for x and identify if it is an extraneous solution.

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2022-06-23T09:35:52+00:00 2 Answers 0

## Answers ( 2 )

1. Answer:

x=21 and solution is not extraneous.

Step-by-step explanation:

Given : Expression [tex]sqrt{(x-5)}+7=11[/tex]

To find : The solution for x and identify if it is an extraneous solution?

Solution :

Expression [tex]sqrt{(x-5)}+7=11[/tex]

[tex]sqrt{(x-5)}=11-7[/tex]

[tex]sqrt{(x-5)}=4[/tex]

Taking square both side,

[tex](sqrt{(x-5)})^2=4^2[/tex]

[tex]x-5=16[/tex]

[tex]x=21[/tex]

Substituting the value back in the equation to find this is an extraneous solution or not.

[tex]sqrt{(21-5)}+7=11[/tex]

[tex]sqrt{(16)}+7=11[/tex]

[tex]4+7=11[/tex]

[tex]11=11[/tex]

This is true.

Which means solution is not an extraneous solution.

2. sqrt(x) – 5+7 =11
combine like terms
-5 +7 = 2

sqrt(x) + 2 = 11
subtract 2 from each side

sqrt(x) = 9

square both sides

x = 9^2

x = 81

an extraneous solution would be an invalid solution

in this case it is not an extraneous solution