F(x) = 4x^2 - 3x + 2kx + 1
What’s the value of k for which the function has two zeros?
Can someone show me step by step?

\( f(x)=4x^2-3x+2kx+1=4x^2+(2k-3)x+1\\\\a=4;\ b=2k-3;\ c=1\\\\function\ has\ two\ zeros\ when\ \Delta=b^2-4ac > 0\\\\\Delta=(2k-3)^2-4\cdot4\cdot1=4k^2-12k+9-16=4k^2-12k-7 > 0\\\\a_k=4;\ b_k=-12;\ c_k=-7\\\\\Delta_k=(-12)^2-4\cdot4\cdot(-7)=144+112=256\\\\k_1=\frac{-b_k-\sqrt{\Delta_k}}{2a_k};\ k_2=\frac{-b_k+\sqrt{\Delta_k}}{2a_k} \)

\( \sqrt{\Delta_k}=\sqrt{256}=16\\\\k_1=\frac{12-16}{2\cdot4}=\frac{-4}{8}=-\frac{1}{2};\ k_2=\frac{12+16}{2\cdot4}=\frac{28}{8}=\frac{7}{2}\\\\a_k=4 > 0\ (up\ parabola\ arms-see\ the\ picture)\\\\Answer: k\in(-\infty;-\frac{1}{2})\ \cup\ (\frac{7}{2};\ \infty) \)


RELATED: