Answer :
[tex]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}[/tex]
[tex]\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)[/tex]
[tex]\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)[/tex]
