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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?

 Dec 28, 2017
 #1
avatar+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

 Dec 28, 2017
edited by MapleTheory  Dec 28, 2017
 #2
avatar+107156 
+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

 

 

cool cool cool

 Dec 29, 2017
 #3
avatar+72 
+1

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

MapleTheory  Dec 29, 2017
 #4
avatar+107156 
+1

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

 

 

cool cool cool

 Dec 29, 2017

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