# Solving for x

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

\$\displaystyle 1 = x(x-6)\$
• Apr 5th 2011, 08:41 AM
Quacky
\$\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?(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.
\$\displaystyle x^2-3x+1x-3=1\$

\$\displaystyle 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