• Nov 5th 2007, 11:04 PM
Local_lunatic
Im stuck on one of my questions in my book, i've got all the others right but this one has really stumped me. Im doing the same steps as the other questions so im pretty sure im missing something. I think im doin something wrong when im expanding each side of the equation and then simplifying.

Use the quadratic formula to solve the equation, rounding you answer to two decimal places.

$\displaystyle 2(x^2-1) = 15(x+3)$
$\displaystyle 2x^2-2 = 15x+45$
$\displaystyle 2x^2-15x-47=0$
$\displaystyle a=2, b=-15, c=-47$

$\displaystyle x= (--15) +- (sqrt)(-15^2 -4(2)(-47)))/4$
$\displaystyle x= (15 +- (sqrt)151)/4$
$\displaystyle x= (15+(sqrt)151)/4 or (15-(sqrt)151)/4$

Does my attempt at writing the question make sense? Thanks :)
$\displaystyle x=6.82 or x=0.68$
• Nov 6th 2007, 12:36 AM
DivideBy0
The discriminant is incorrect, the rest of the problem is correct.

$\displaystyle 2x^2-15x-47=0$

$\displaystyle x=\frac{-(-15)\pm \sqrt{(-15)^2-4(2)(-47)}}{2(2)}$

$\displaystyle =\frac{15 \pm \sqrt{225+376}}{4}$

$\displaystyle =\frac{15 \pm \sqrt{601}}{4}$
• Nov 6th 2007, 03:19 AM
topsquark
Quote:

Originally Posted by Local_lunatic
$\displaystyle x=6.82 or x=0.68$

Unless you have a specific reason, you should leave your answer in exact form, not in decimals.

-Dan