Equations Involving Radicals Question

[HerroFish]

Solve for:
√(x-7) / √(x) -2 = √2

My attempt at a solution:
I solved for x and it comes out to:

0 = x^2 - 64x +225

and then i plugged it into the quadratic formula:

[-(-64)±√((64)^2-4(1)(225))]/2

and my answer comes out to be:

32±√799

although the answer on the back of the package is 9 and 25.

kind of confused on how to attempt this question now and not sure what I'm doing wrong...

And thanks in advance for the help!

 

[Dick]

Solve for:
√(x-7) / √(x) -2 = √2

My attempt at a solution:
I solved for x and it comes out to:

0 = x^2 - 64x +225

and then i plugged it into the quadratic formula:

[-(-64)±√((64)^2-4(1)(225))]/2

and my answer comes out to be:

32±√799

although the answer on the back of the package is 9 and 25.

kind of confused on how to attempt this question now and not sure what I'm doing wrong...

And thanks in advance for the help!

0 = x^2 - 64x +225 isn't the right quadratic. Check it again or show how you got it.

 

[HerroFish]

0 = x^2 - 64x +225 isn't the right quadratic. Check it again or show how you got it.

(√(x-7)) / (√(x) - 2) = √2

√(x-7) = √(2) (√(x) - 2)

x-7 = (√2x - 2√2) ^2 <------- square both sides

x-7 = (√2x)^2 - 2(√2x)(2√2) + (√8)^2

x-7 = 2x - 8√x + 8

0 = x - 8√x +15

0 = x^2 - 64x + 225 <----- squared both sides to get rid of radical

and plug into quadratic eqn. :P

 

[SammyS]

Solve for:
√(x-7) / √(x) -2 = √2

Is the problem your working on,

\displaystyle \frac{\sqrt{x-7\,}}{\sqrt{x\,}-2}=\sqrt{2} \ ?

Or is it \displaystyle \frac{\sqrt{x-7\,}}{\sqrt{x\,}}-2=\sqrt{2} \ ? which is literally what you wrote.

 

[HerroFish]

Is the problem your working on,

\displaystyle \frac{\sqrt{x-7\,}}{\sqrt{x\,}-2}=\sqrt{2} \ ?

Or is it \displaystyle \frac{\sqrt{x-7\,}}{\sqrt{x\,}}-2=\sqrt{2} \ ? which is literally what you wrote.

the first one
Sorry i don't really know how to use the square root sign, pretty new to this

 

[Dick]

(√(x-7)) / (√(x) - 2) = √2

√(x-7) = √(2) (√(x) - 2)

x-7 = (√2x - 2√2) ^2 <------- square both sides

x-7 = (√2x)^2 - 2(√2x)(2√2) + (√8)^2

x-7 = 2x - 8√x + 8

0 = x - 8√x +15

0 = x^2 - 64x + 225 <----- squared both sides to get rid of radical

and plug into quadratic eqn. :P

You were doing fine till you tried to square both sides of 0 = x - 8√x +15. Squaring both sides won't get rid of the radical if you do the algebra right. Best to write it as 8√x = x+15 first. Now square both sides.

 

[HerroFish]

You were doing fine till you tried to square both sides of 0 = x - 8√x +15. Squaring both sides won't get rid of the radical if you do the algebra right. Best to write it as 8√x = x+15 first. Now square both sides.

OHHHHH That make sense!

thanks!

 

[SammyS]

the first one
Sorry i don't really know how to use the square root sign, pretty new to this

It's more a problem of leaving out or of misplacing parentheses.

Writing , √(x-7) / ( √(x) -2 ) = √2 would be OK .

 

[Ray Vickson]

the first one
Sorry i don't really know how to use the square root sign, pretty new to this

Square toot signs are not the issue; brackets are what you missed. Everything would have been clear if you had written (√(x-7)) / (√(x) - 2) = √2 . The point is that writing something like A/B-C means (A/B) - C if you read it using standard priority rules for mathematical expressions. If you want A/(B-C), you need brackets.

RGV