1. ## Solving for x

How do I find x for

$\displaystyle 1 = x(x-6)$

2. $\displaystyle x(x-6)=1$
Expand the brackets by multiplying everything inside the bracket with everything outside the bracket term by term to obtain:

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

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

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?

3. 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?

4. 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.
$\displaystyle x^2-3x+1x-3=1$

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

etc...

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