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Solve for x.

√x+5= √x+ 1

no solution

4

−4

2

I think it would be 4? I am not sure.

jjennylove Oct 24, 2018

#3**+2 **

Solve for x:

sqrt(x + 5) = sqrt(x) + 1

Raise both sides to the power of two:

x + 5 = (sqrt(x) + 1)^2

Subtract (sqrt(x) + 1)^2 from both sides:

5 - (sqrt(x) + 1)^2 + x = 0

5 - (sqrt(x) + 1)^2 + x = 4 - 2 sqrt(x):

4 - 2 sqrt(x) = 0

Subtract 4 from both sides:

-2 sqrt(x) = -4

Divide both sides by -2:

sqrt(x) = 2

Raise both sides to the power of two:

**x = 4**

Guest Oct 24, 2018

#1**+1 **

As you have it written....no solution

Think about this....whatever √ x is on one side it is the same thing on the other side

But adding 5 to this value on one side will always be 3 more than adding 2 to it on the other side

So.... the left side > right side for any x value ≥ 0

CPhill Oct 24, 2018

#2**0 **

This is how it looks would it still be no soultion ? i am kinda confused.

the square root goes only goes across x+5 and the square root just goes on x for the second equation

jjennylove
Oct 24, 2018

#3**+2 **

Best Answer

Solve for x:

sqrt(x + 5) = sqrt(x) + 1

Raise both sides to the power of two:

x + 5 = (sqrt(x) + 1)^2

Subtract (sqrt(x) + 1)^2 from both sides:

5 - (sqrt(x) + 1)^2 + x = 0

5 - (sqrt(x) + 1)^2 + x = 4 - 2 sqrt(x):

4 - 2 sqrt(x) = 0

Subtract 4 from both sides:

-2 sqrt(x) = -4

Divide both sides by -2:

sqrt(x) = 2

Raise both sides to the power of two:

**x = 4**

Guest Oct 24, 2018