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3 years ago

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

2 answers:

8 0

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

ddd [48]3 years ago

5 0

Answer:

x=21 and solution is not extraneous.

Step-by-step explanation:

Given : Expression $\sqrt{(x-5)}+7=11$

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

Solution :

Expression $\sqrt{(x-5)}+7=11$

$\sqrt{(x-5)}=11-7$

$\sqrt{(x-5)}=4$

Taking square both side,

$(\sqrt{(x-5)})^2=4^2$

$x-5=16$

$x=21$

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

$\sqrt{(21-5)}+7=11$

$\sqrt{(16)}+7=11$

$4+7=11$

$11=11$

This is true.

Which means solution is not an extraneous solution.

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