If the two roots of the quadratic \( 4x^2+7x+k\) are \( \frac{-7\pm i\sqrt{15}}{8}\), what is k?

Guest Dec 28, 2017

#1**+1 **

***Since I'm not sure if that "i" before ****the ****\(\sqrt{15}\)**** ,**** I'****m**** going to assume ****it's**** not there. If another user knows what to do with the "i" then you should follow their answer.***

---------------------------------------------------------------------------------------------------------------------------------------

Given the quadratic formula: \(x = {-b \pm \sqrt{b^2-4ak} \over 2a}\) we already know what the variable a and b are 4 and 7, respectively.

Now, we plug in the knowns into the formula, and we get: \(x = {-7 \pm \sqrt{7^2-4(4)(k)} \over 2(4)}\)

As a result, you get: \(x = {-7 \pm \sqrt{49-16k} \over 8}\)

Now, since we know that the number inside the square root is 15, to solve for k you can do this:

49-16k = 15

-16k = 15-49

-16k = -34

k = -34 / -16

k = 2.125

**Therefore, k equals to 2.125**

MapleTheory
Dec 28, 2017