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How do you find x in this equation? sqrt(x+11) = sqrt(x)+1

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jabraeq | Student, Undergraduate | (Level 2) eNoter

Posted June 16, 2010 at 4:34 AM via web

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How do you find x in this equation? sqrt(x+11) = sqrt(x)+1

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brianevery | High School Teacher | (Level 2) Adjunct Educator

Posted June 16, 2010 at 6:18 AM (Answer #1)

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For this problem, you are going to square both sides of the equation:

(sqrt(x+11))^2=(sqrt(x)+1)^2

Which gives us:

x+11 = x+2sqrt(x) +1

We will then subtract both x and 1 from both sides of the equation giving us:

x+11-x-1 = x+ 2sqrt(x)+1-x-1

Then combine like terms:

10 = 2 sqrt(x)

Divide both sides by 2:

5=sqrt(x)

And square both sides again:

25 = x

Then check the answer in the original equation to see if it is extraneous:

sqrt(25+11) = sqrt(25)+1

sqrt (36) = 5+1

6=6

It checks!

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted June 16, 2010 at 6:28 AM (Answer #2)

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sqrt(x+11) = sqrt(x)+1

Square both sides:

x+11 = x+2sqrt(x) +1

Reduce similar:

10=2sqrt(x)

Divide by 5:

==> sqrt(x) = 5

==> x= 25

Now to check your answer:

sqrt(25+11) = sqrt(25)+1

sqrt(36) = 5+1

6= 6

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tonys538 | Student, Undergraduate | TA | (Level 1) Valedictorian

Posted January 3, 2015 at 1:33 PM (Answer #3)

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The solution of the equation `sqrt(x+11) = sqrt(x)+1` has to be determined.

Taking the square of both sides of the equation does not affect the equation.

`(sqrt (x+11))^2 = (sqrt x + 1)^2`

`(x + 11) = x + 1 + 2*sqrt x`

`2*sqrt x = 10`

`sqrt x = 5`

x = 25

The solution of the equation `sqrt(x+11) = sqrt(x)+1` is x = 25

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