Quadratic Formula

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November 5th 2007, 11:04 PM
Local_lunatic

Quadratic Formula

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.

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

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

Does my attempt at writing the question make sense? Thanks :)
$x=6.82 or x=0.68$
November 6th 2007, 12:36 AM
DivideBy0

The discriminant is incorrect, the rest of the problem is correct.

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

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

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

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

Quote:

Originally Posted by Local_lunatic

$x=6.82 or x=0.68$

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

-Dan

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