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

rn50nurse

New member
Joined
Aug 5, 2007
Messages
22
Hi-- can anyone help me solve this??

3+ sq rootx^2-8x=0
thanks!
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
18,133
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
Joined
Aug 5, 2007
Messages
22
yes!

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

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
18,133
So now tell me

sqrt(x^2) = ???
 

rn50nurse

New member
Joined
Aug 5, 2007
Messages
22
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
Joined
Sep 28, 2005
Messages
7,216
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
Joined
Dec 17, 2005
Messages
2,128
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
Joined
Aug 5, 2007
Messages
22
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
Joined
Feb 17, 2004
Messages
1,442
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
Joined
Jun 18, 2007
Messages
18,133
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).
 
Top