If the two roots of the quadratic 4x^2 + 7x + k are (-7 +/- i*sqrt(79))/8, what is k?
+2666 By the quadratic formula, the roots are: \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\).
Substituting what we know, we have: \(x = {-7 \pm \sqrt{49-16c} \over 8}\)
This means that the discriminate must equal -79, giving us the equation: \(49 - 16c = -79\)
Subtracting 49 from both sides, we have: \(-16c = -128\), meaning \(k = \color{brown}\boxed{8}\)
+2666 By the quadratic formula, the roots are: \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\).
Substituting what we know, we have: \(x = {-7 \pm \sqrt{49-16c} \over 8}\)
This means that the discriminate must equal -79, giving us the equation: \(49 - 16c = -79\)
Subtracting 49 from both sides, we have: \(-16c = -128\), meaning \(k = \color{brown}\boxed{8}\)
