# Solving for x

• Apr 5th 2011, 08:32 AM
Devi09
Solving for x
How do I find x for

$1 = x(x-6)$
• Apr 5th 2011, 08:41 AM
Quacky
$x(x-6)=1$
Expand the brackets by multiplying everything inside the bracket with everything outside the bracket term by term to obtain:

$x^2-6x=1$ Then rearrange:

$x^2-6x-1=0$

Which is just a standard solvable quadratic, but it won't factorize, so you'll either have to use the formula or complete the square. Can you remember how to do that?(Wink)
• Apr 5th 2011, 08:45 AM
Devi09
Yeh. Plug in the values and you get x= 6.1623 and x= -0.16228

And if you were to solve for 1 = (x-3)(x+1)

you just expand, set the equation to 0 and find the x's right?
• Apr 5th 2011, 08:48 AM
Quacky
Quote:

Originally Posted by Devi09
Yeh. Plug in the values and you get x= 6.1623 and x= -0.16228

And if you were to solve for 1 = (x-3)(x+1)

you just expand and find the x's right?

Yeah, it's a similar approach.
$x^2-3x+1x-3=1$

$x^2-2x-4=0$

etc...
• Apr 5th 2011, 07:15 PM
Prove It
Quote:

Originally Posted by Devi09
Yeh. Plug in the values and you get x= 6.1623 and x= -0.16228