# Thread: x = 9 times the square root of x

1. ## x = 9 times the square root of x

Can someone help me with this problem? I do not know how to work it...

x = 9 (the square root of x)

2. x = 9*sqrt(x) => x^2 = 81x

which gives x^2 - 81x = 0

or

x(x - 81) = 0

x= 0, 81

3. The square root of x = 9

If that's what you're asking for then it's simply 9^2 which is 81.
So the square root of 81 = 9.

If that's not what you're asking for, then sorry I dont full understand what you're asking.

x = 9*sqrt(x) => x^2 = 81x

which gives x^2 - 81x = 0

or

x(x - 81) = 0

x= 0, 81
Ahh... this is it. Thanks for the explanation.

5. Originally Posted by TMU
The square root of x = 9

If that's what you're asking for then it's simply 9^2 which is 81.
So the square root of 81 = 9.

If that's not what you're asking for, then sorry I dont full understand what you're asking.
The reason you missed the x = 0 solution is that you divided both sides by $\sqrt{x}$ which is an invalid procedure when x = 0.

Tip: When you are dividing both sides of an equation always remember to set that factor equal to 0 to see if there might be another solution in there.

-Dan