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

[Guest]

factoring quadractic equations:

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

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

 

[ELIZABETH J.]

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.

 

[daon]

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.