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.