# Math Help - Simultaneous Equations with square root

1. ## Simultaneous Equations with square root

Hi, I'm trying to find the solution to $\sqrt{x}=x-6$ but i can't for the life of me figure out how to do it, it's the square root that has me stumped. I've managed to get it into a form of $x^2 +x+-36$ and found the roots of approximately 5.5 and -6.5 which i then substituted into the y=mx+c form equations but still couldnt get the right answer of 4,2.
So I've returned again seeking help, anyone care to take a stab?

2. Originally Posted by mattty
Hi, I'm trying to find the solution to $\sqrt{x}=x-6$ but i can't for the life of me figure out how to do it, it's the square root that has me stumped. I've managed to get it into a form of $x^2 +x+-36$ and found the roots of approximately 5.5 and -6.5 which i then substituted into the y=mx+c form equations but still couldnt get the right answer of 4,2.
So I've returned again seeking help, anyone care to take a stab?
$x=(\sqrt{x})^2$
So the equation is now $y=y^2-6$, where $y=\sqrt{x}$

now can you solve it ? don't forget that y is positive since the square root of a real number is always positive.

3. to be honest... no, i still can't. I suppose I should have put in more information in my original post, but that was one of the steps i went through to get to $x^2 +x-36$ I suppose the real problem I'm having is when i get $x+\sqrt{x}=-6$ because i can't figure out how to bring it down to a single x term.

4. Originally Posted by mattty
to be honest... no, i still can't. I suppose I should have put in more information in my original post, but that was one of the steps i went through to get to $x^2 +x-36$ I suppose the real problem I'm having is when i get $x+\sqrt{x}=-6$ because i can't figure out how to bring it down to a single x term.
Following what I have said, put them all on the same side :
you get : $y^2-y-6=0$ (I definitely don't know how you get yours)

Now this has two roots (which you get by using whatever method you want) : 3 and -2
Since y>0, y=3
So x=+ or - sqrt(3)

5. hmmm, that still doesn't satisfy the solution of 4,2 ( $x=4, y=2$) that I gave in my original post. According to the graph on my calculator, 4,2 is the point where the 2 equations meet. Was I asking for the wrong thing perhaps?

6. Originally Posted by Moo
Following what I have said, put them all on the same side :
you get : $y^2-y-6=0$ (I definitely don't know how you get yours)

Now this has two roots (which you get by using whatever method you want) : 3 and -2
Since y>0, y=3
So x=+ or - sqrt(3)
I made a very stupid mistake.........................

So we have y^2-y-6=0
This means that y=3 or y=-2
But x=y² (and not the contrary)
So x=9 or 4.

If you substitute in $\sqrt{x}=x-6$, 9 will work, but not 4, because $\sqrt{4} \neq 4-6$
So 9 is a solution

I didn't understand this was for finding the intersection of two curves... So you want to find the intersection between y=sqrt(x) and y=x-6 ?
9,3 is the correct answer. And I checked it on a graph...

It would have been nice to put the whole problem ^^' to avoid confusion. And I'm sorry for the one I created.

7. Edit: i'll start a new thread tomorrow, thanks for the help, misguided as it was, Moo