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If the two roots of the quadratic 4x^2 + 7x + k are (-7 +/- i*sqrt(79))/8, what is k?

Jun 21, 2022

#1
+2602
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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 $$k = \color{brown}\boxed{8}$$

Jun 21, 2022

#1
+2602
0

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 $$k = \color{brown}\boxed{8}$$

BuilderBoi Jun 21, 2022