# Determine values of k for which function has two zeros.

• Apr 2nd 2010, 08:48 AM
Determine values of k for which function has two zeros.
Determine the values of k for which the function $\displaystyle f(x) = 4x^2 - 3x + 2kx +1$ has two zeros. Check the values in the original equation.

$\displaystyle a = 4, b= (-3 + 2k) , c = 1$
$\displaystyle b^2 - 4ac > 0$
$\displaystyle (-3 + 2k)^2 - 4(4)(1) > 0$
$\displaystyle 9 + 4k^2 - 16 > 0$
$\displaystyle 4k^2 > -9 + 16$
$\displaystyle \frac{4k^2 > 7}{4}$
$\displaystyle k^2 > \frac{7}{4}$
$\displaystyle \sqrt{k^2} > \sqrt{\frac{7}{4}}$
$\displaystyle k > \pm \frac{\sqrt{7}}{2}$

The textbook answer to this question is "$\displaystyle x < -0.5$ or $\displaystyle x > 3.5$". Where did I go wrong in solving this? Was it the part where I let $\displaystyle b = (-3 + 2k)$? I wasn't sure about that as there were two values of x in the standard form.
• Apr 2nd 2010, 09:04 AM
alexmahone
$\displaystyle a=4, b=(-3 + 2k), c=1$

$\displaystyle b^2-4ac>0$

$\displaystyle (-3+2k)^2-4(4)(1)>0$

$\displaystyle 9 + 4k^2-12k-16 > 0$

$\displaystyle 4k^2-12k-7>0$

$\displaystyle 4k^2+2k-14k-7>0$

$\displaystyle 2k(2k+1)-7(2k+1)>0$

$\displaystyle (2k+1)(2k-7)>0$

$\displaystyle x<-0.5$ or $\displaystyle x>3.5$
• Apr 2nd 2010, 09:18 AM
Tyvm for the reply. Didn't even realize that I squared the bracketed numbers wrong. Just one question though. How did you get $\displaystyle (+2k - 14k)$ from $\displaystyle -12k$?
• Apr 2nd 2010, 10:29 AM
jayAndy
Quote:

Tyvm for the reply. Didn't even realize that I squared the bracketed numbers wrong. Just one question though. How did you get $\displaystyle (+2k - 14k)$ from $\displaystyle -12k$?