# Finding x-intercepts

• Nov 12th 2012, 12:16 PM
mrosa
Finding x-intercepts
h(x)= 2x2-5x-4

The back of the books tells me that the answer is (5 - the square root of 57 over 4. ,0) , (5 + the square root of 57 over 4. ,0) but I can't figure out how that was gotten. In the section above, we were working on completing the square so am I supposed to do the same and then find the intercepts by replacing y with 0?

• Nov 12th 2012, 12:26 PM
MarkFL
Re: Finding x-intercepts
Yes, equate the function to zero to find its roots, which are the x-intercepts. Can you show your work for completing the square so we may find where the error(s) may be so you know what you are doing wrong? Your book has the correct roots.
• Nov 12th 2012, 12:29 PM
skeeter
Re: Finding x-intercepts
$2x^2-5x-4 = 0$

$x^2 - \frac{5}{2}x = 2$

$x^2 - \frac{5}{2}x + \frac{25}{16} = 2 + \frac{25}{16}$

$\left(x - \frac{5}{4}\right)^2 = \frac{57}{16}$

$x - \frac{5}{4} = \pm \frac{\sqrt{57}}{4}$

$x = \frac{5 \pm \sqrt{57}}{4}$
• Nov 13th 2012, 10:06 AM
mrosa
Re: Finding x-intercepts
Thank you,