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# ???HELP???

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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
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#1
+72
+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.*

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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
edited by MapleTheory  Dec 28, 2017
#2
+80905
+2

I think the "i" is supposed to be there, MapleTheory

So....

The quantity  under the radical  must be  = -15

So......we can solve this...

7^2  - 4(4)k  = - 15

49  -  16k   =   -15

-16k  =  - 64

k  =  4

CPhill  Dec 29, 2017
#3
+72
+1

ah, okay. I was unaware of what the "i" meant, thanks!

MapleTheory  Dec 29, 2017
#4
+80905
+1

No prob....I still gave you some points....!!!

CPhill  Dec 29, 2017

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