Solving this simple function

**fattydq** · March 9th 2009, 01:31 PM

X^2-5x-3=0?

I honestly have no idea how to do this, perhaps x^2-5x=3, but then I'm still stuck, help!

**JeWiSh** · March 9th 2009, 01:48 PM

Originally Posted by fattydq

X^2-5x-3=0?

I honestly have no idea how to do this, perhaps x^2-5x=3, but then I'm still stuck, help!

$x^2-5x-3=0$
It doesn't factorise easily, do you know the quadratic formula?
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ So in this case a=1 b=-5 and c=-3

**sammy28** · March 9th 2009, 01:52 PM

leave the equation equal to 0 to solve.

since it can't be factorized you can solve either by

$\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$

where a, b, c are the coefficients of the terms, or you can solve using 'completing the square'

$(x - \frac{5}{2})^2 - \frac{25}{4} - 3$

$(x - \frac{5}{2})^2 = \frac{37}{4}$

$x = \frac{5}{2} \pm \sqrt{\frac{37}{4}}$

**fattydq** · March 9th 2009, 01:52 PM

Originally Posted by JeWiSh

$x^2-5x-3=0$
It doesn't factorise easily, do you know the quadratic formula?
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ So in this case a=1 b=-5 and c=-3

Oh yeah, I did already know that but for some reason my mind just blanked when I saw it. Thanks a bunch!

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