# need help ASAP with 3+ sq rootx^2-8x=0

#### rn50nurse

##### New member
Hi-- can anyone help me solve this??

3+ sq rootx^2-8x=0
thanks!

#### Subhotosh Khan

##### Super Moderator
Staff member
Re: need help ASAP

rn50nurse said:
Hi-- can anyone help me solve this??

3+ sq rootx^2-8x=0
thanks!
Is the problem

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

or some other thing.

If it is as I wrote it -

what is sqrt(x^2) = ??

#### rn50nurse

##### New member
yes!

so sorry- it is hard to put these down correctly- i beleive you rewrote it correctly! thanks

Staff member
So now tell me

sqrt(x^2) = ???

#### rn50nurse

##### New member
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???

#### galactus

##### Super Moderator
Staff member
Oh, see what happens when you don't properly use parentheses?. I, like SK, thought that was $$\displaystyle \L\\3+\sqrt{x^{2}}-8x=0$$

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.

#### Mrspi

##### Senior Member
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???
Ok...is this correct?

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.

#### rn50nurse

##### New member
Awesome--thank you! I just finished up doing just that and that is what I got also-- the post prior toyours said something about La Tex? how do I get that to help is it the tab above here that says TeX? Thanks for your help!!!

#### Denis

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

#### Subhotosh Khan

##### Super Moderator
Staff member
However, as MrsPi indicated - check the answer.

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

That means the problem does not have a solution.

Remember, [square root] of anything, is always positive (in real domain).