05-12-2006, 01:43 PM

factoring quadractic equations:

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

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

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

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

View Full Version : help solve -2x^2 + 11x - 5 = 0

05-12-2006, 01:43 PM

factoring quadractic equations:

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

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

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

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

ELIZABETH J.

05-12-2006, 02:02 PM

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. :D

Hope it helps. :D

daon

05-12-2006, 02:39 PM

Divide through by -2..

-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:

x^2 - \frac{11}{2} x + \frac{5}{2} + \frac{121}{16} - \frac{121}{16} = 0

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

5) So, 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 (x^2 - \frac{11}{4})^2 - \frac{81}{16} = 0 solve for x:

(x - \frac{11}{4})^2 = \frac{81}{16}

x - \frac{11}{4} = \, +/- \, \sqrt{\frac{81}{16}}

x = \frac{11}{4} \, +/- \, \frac{\sqrt{81}}{4}

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. :)

-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:

x^2 - \frac{11}{2} x + \frac{5}{2} + \frac{121}{16} - \frac{121}{16} = 0

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

5) So, 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 (x^2 - \frac{11}{4})^2 - \frac{81}{16} = 0 solve for x:

(x - \frac{11}{4})^2 = \frac{81}{16}

x - \frac{11}{4} = \, +/- \, \sqrt{\frac{81}{16}}

x = \frac{11}{4} \, +/- \, \frac{\sqrt{81}}{4}

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. :)

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