$\displaystyle \mbox{if}\ a^2x^2+6abx+ac+8b^2=0\ \mbox{have equal roots then,prove}\ ac(x+1)^2 =4b^2x\ \mbox{have equal roots.}$

Results 1 to 5 of 5

- May 16th 2012, 10:15 PM #1

- May 16th 2012, 11:08 PM #2

- May 17th 2012, 04:48 AM #3

- May 17th 2012, 05:00 AM #4

- May 17th 2012, 05:24 AM #5

- Joined
- Dec 2007
- From
- Ottawa, Canada
- Posts
- 3,184
- Thanks
- 80

## Re: Algeb

$\displaystyle \frac{-6ab \pm \sqrt{36a^2b^2 - 4a^2(ac+8b^2)}}{2a^2}$

36 a^2 b^2 = 4 a^2(ac + 8 b^2); simplifies to b^2 = ac

Now continue with other equation...