1. ## Rearranging a quadratic equation

y=(4-x)(2+x)

Can this be rearranged to make an equation x=something without x being on the rhs?

2. Originally Posted by Stuck Man
y=(4-x)(2+x)

without x being on the rhs?
y=(4-x)(2+x)

y=8+4x-2x-x^2

y=8-2x-x^2

x^2+2x+y=8

3. I want the lhs to be x.

4. ## hmm

x(x+2) = 8-y

5. No the lhs has to be x alone. I don't think it can be done.

6. ## right

Originally Posted by Stuck Man
No the lhs has to be x alone. I don't think it can be done.
if it has to be x alone i am sure it cant be done

7. ## yess

8. x= sqrt(9-y)+1

I moved the graph left by 1 to have the axis of symmetry at the y-axis. Then I rearranged it and added 1.

y=-x^2+9
x=sqrt(9-y)
x=sqrt(9-y)+1

9. Originally Posted by Stuck Man
y=(4-x)(2+x)

Can this be rearranged to make an equation x=something without x being on the rhs?
$y = 8+2x-x^2$

$x^2-2x+8-y = 0$

You can use the quadratic formula to find x

$x = 1 \pm \sqrt{y-7}$

10. Originally Posted by e^(i*pi)
$y = 8+2x-x^2$

$x^2-2x+8-y = 0$

You can use the quadratic formula to find x

$x = 1 \pm \sqrt{y-7}$
ya this can b done...

11. There are two signs wrong after rearranging.

After using the quadratic formula I have come up with a different but equivalent equation to my earlier one.

I can also make it equal to my earlier equation using some interesting algebra that I've thought of.

12. Originally Posted by Stuck Man
There are two signs wrong after rearranging.

After using the quadratic formula I have come up with a different but equivalent equation to my earlier one.

I can also make it equal to my earlier equation using some interesting algebra that I've thought of.
i guess the bfore eq was a bit complex....maybe fr me

13. I've attached a scan of my work. I was planning to do integration against the y-axis but it happens to be more work.