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

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

Determine the values of k for which the function $f(x) = 4x^2 - 3x + 2kx +1$ has two zeros. Check the values in the original equation.

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

The textbook answer to this question is " $x < -0.5$ or $x > 3.5$". Where did I go wrong in solving this? Was it the part where I let $b = (-3 + 2k)$? I wasn't sure about that as there were two values of x in the standard form.

2. $a=4, b=(-3 + 2k), c=1$

$b^2-4ac>0$

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

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

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

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

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

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

$x<-0.5$ or $x>3.5$

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

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