help solve -2x^2 + 11x - 5 = 0

factoring quadractic equations:

-2x^2+11x-5=0

do i divide by a negative one or what number do i use?

The first thing you have to do is factor so once you have factor the equation set the factors to equal zero.You have to solve each factor


Hope it helps.

Divide through by -2..

\(\displaystyle -2x^2 + 11x - 5 = 0 \Rightarrow x^2 - \frac{11}{2} x + \frac{5}{2} = 0 \\\)

Now complete the square:

1) divide -11/2 by 2 you get -11/4

2) square result from number 1.. 121/16

3) Add and subtract 121/16 from the original equation:
\(\displaystyle x^2 - \frac{11}{2} x + \frac{5}{2} + \frac{121}{16} - \frac{121}{16} = 0\)

4) As a direct result, \(\displaystyle x^2 - \frac{11}{2} x + \frac{121}{16}\) factors into: \(\displaystyle (x - \frac{11}{4})^2\)

5) So, \(\displaystyle x^2 - \frac{11}{2} x + \frac{5}{2} + \frac{121}{16} - \frac{121}{16} = (x - \frac{11}{4})^2 + \frac{5}{2} - \frac{121}{16} = (x - \frac{11}{4})^2 + \frac{40}{16} - \frac{121}{16} = (x - \frac{11}{4})^2 - \frac{81}{16} = 0 \\\)

6) Finally, given \(\displaystyle (x^2 - \frac{11}{4})^2 - \frac{81}{16} = 0\) solve for x:
\(\displaystyle (x - \frac{11}{4})^2 = \frac{81}{16}\)
\(\displaystyle x - \frac{11}{4} = \, +/- \, \sqrt{\frac{81}{16}}\)
\(\displaystyle x = \frac{11}{4} \, +/- \, \frac{\sqrt{81}}{4}\)
\(\displaystyle x = \frac{11\,\, +/- \,\, 9}{4}\)

x = 5 or x = 1/2

You could also use the quadratic equation, which in many cases will be easier.