- Thread starter rn50nurse
- Start date

- Joined
- Jun 18, 2007

- Messages
- 18,133

Is the problemrn50nurse said:Hi-- can anyone help me solve this??

3+ sq rootx^2-8x=0

thanks!

3 + (sq rootx^2) - 8x = 0

or some other thing.

If it is as I wrote it -

what is sqrt(x^2) = ??

- Joined
- Jun 18, 2007

- Messages
- 18,133

So now tell me

sqrt(x^2) = ???

sqrt(x^2) = ???

Instead it's \(\displaystyle \L\\3+\sqrt{x^{2}-8x}=0?\)

Take a shot at learning some LaTex. It makes things so much easier to read.

Anyway, subtract 3 from both sides and then square both sides to eliminate the radical. After you shed the radical, you can solve for x easily.

Ok...is this correct?rn50nurse said:the problem states- 3+ outside the sq root symbol and inside the sq root symbol is x^2-8 then =0 solve for x. does that make sense???

3 + sqrt(x<SUP>2</SUP> - 8x) = 0

If so, I'd subtract 3 from both sides of the equation to isolate the radical expression:

sqrt(x<SUP>2</SUP> - 8x) = -3

Square both sides to eliminate the radical:

[sqrt(x<SUP>2</SUP> - 8x)]<SUP>2</SUP> = (-3)<SUP>2</SUP>

x<SUP>2</SUP> - 8x = 9

x<SUP>2</SUP> - 8x - 9 = 0

Can you factor the left side of the equation?

(x - 9)(x + 1) = 0

Set each factor equal to 0, and solve for x.

Be sure to check your solutions in the original equation; squaring both sides of an equation MAY produce exttraneous solutions which do NOT check in the original equation.

Next time, use "sqrt" and BRACKETS around what's being square rooted:rn50nurse said:3+ sq rootx^2-8x=0

3 + sqrt(x^2 - 8x) = 0 : got that?

move the 3 on other side:

sqrt(x^2 - 8x) = -3

square both sides:

x^2 - 8x = 9

move the 9 to left side:

x^2 - 8x - 9 = 0

factor the above:

(x - 9)(x + 1) = 0

so x - 9 = 0 or x + 1 = 0; so x = 9 or x = -1

nurse, if you're unable to follow that, you need a doctor :lol:

- Joined
- Jun 18, 2007

- Messages
- 18,133

As you have corrected the posted problem - now none of the solutions are

That means the problem does not have a solution.

Remember, [square root] of anything, is