Completing the square.

• Apr 14th 2008, 10:59 AM
Mymble
Completing the square.
I need to solve this equation by completing the square:

2x^2 + x - 8 = 0

In my textbook the answer is x = 1.77 or x = -2.27

I've tried to do it but can't get the right answer...
2x^2 + x - 8 = 0
x^2 + 0.5x - 4 = 0
x^2 + 0.5x = 4
x^2 + 0.5x + 0.0625 = 4.0625
(x + 0.0625)^2 = 4.0625
x + 0.0625 = (+/-)sqrt4.0625
x = - 0.0625 + sqrt4.0625 or x = - 00625 - sqrt4.0625
x = 1.95... or x = -2.07...

Thanks.
• Apr 14th 2008, 11:16 AM
Aryth
Ok, so we have the equation:

$2x^2 + x - 8 = 0$

Step 1: Move the term without an x to the other side of the equation:

$2x^2 + x = 8$

Step 2: get the $x^2$ term by itself by dividing by its coefficient:

$x^2 + \frac{1}{2}x = 4$

Step 3: Set aside the coefficient of the x-term and divide it in half:

$\frac{\frac{1}{2}}{2} = \frac{1}{4}$

$\left(\frac{1}{4}\right)^2 = \frac{1}{16}$

Step 5: Add the square to both sides:

$x^2 + \frac{1}{2} + \frac{1}{16} = \frac{65}{16}$

Step 6: Set up the factored form:

$\left(x + \frac{1}{4}\right)^2 = \frac{65}{16}$

Step 7: Take the square root of both sides:

$x + \frac{1}{4} = \pm \frac{\sqrt{65}}{4}$

Step 8: Separate the plus and minus:

$x + \frac{1}{4} = \frac{\sqrt{65}}{4}$

$x + \frac{1}{4} = -\frac{\sqrt{65}}{4}$

Step 9: Solve each equation:

$x = \frac{\sqrt{65}-1}{4}$

$x = -\frac{\sqrt{65}+1}{4}$

Put the fractions in the calculator and you get:

$x \approx 1.7655 \approx 1.77$

$x \approx 2.2655 \approx 2.27$

And there you go.
• Apr 14th 2008, 11:30 AM
Mymble
Thankyou very much.

I think I went wrong when I should have had (x + 1/4)^2 but I had (x + 1/16)^2.