• Jan 18th 2011, 11:06 AM
Kromletch
1/2x^2+3/4x-1=0

-3/4(+-)SQRT((3/4^2)-4(1/2)(-1) / 2(1/2)

-3/4(+-)SQRT(9/16+2) / 1

This is as far as I can get I do not see any errors yet the answer is supposed to be

-3+-SQRT(41) /4

I am clueless as to where the 41 comes from
• Jan 18th 2011, 11:16 AM
Ackbeet
The original problem:

$\displaystyle \dfrac{1}{2}\,x^{2}+\dfrac{3}{4}\,x-1=0.$

$\displaystyle x=\dfrac{-3/4\pm\sqrt{9/16-4(1/2)(-1)}}{1}=-3/4\pm\sqrt{9/16+2}= -3/4\pm\sqrt{(9+32)/16}=\dfrac{-3\pm\sqrt{41}}{4}.$

That's what I get. WolframAlpha gives the same.

• Jan 18th 2011, 11:29 AM
Kromletch
My apologies the books answer is yours as well i typed it incorrectly. How do you get (9 +32) / 16 out of 9/16+2
• Jan 18th 2011, 11:31 AM
Ackbeet
You have

$\displaystyle \dfrac{9}{16}+2=\dfrac{9}{16}+\dfrac{32}{16}=\dfra c{9+32}{16}=\dfrac{41}{16}.$

This is the usual addition of fractions: get a common denominator and proceed.
• Jan 18th 2011, 11:40 AM
Quacky
Obviously it isn't necessary, but it would be easier to calculate if you multiplied the equation by four.

$\displaystyle 2x^2 + 3x - 4 = 0$ is so much easier to substitute into the formula, and it will probably save time as it's so much simpler to follow through.
• Jan 18th 2011, 11:40 AM
Wilmer
Quote:

Originally Posted by Kromletch
1/2x^2+3/4x-1=0
This is as far as I can get I do not see any errors yet the answer is supposed to be
-3+-SQRT(41) /4

To start, that should be: [-3 +- SQRT(41)] /4 ; the extra brackets are IMPORTANT.

GOOD IDEA to always get rid of fractions before doing anything else;
so multiply your equation by 4: 2x^2 + 3x - 4 = 0 : get the picture?
Now use the quadratic and you'll have no problems (Wink)