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AVprozaik [17]
3 years ago
6

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

Mathematics
2 answers:
insens350 [35]3 years ago
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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