SOLUTION: I am supposed to find k in the following quadratic equation: (k+3)x^2 + 2kx + 4 = 0. I have tried using the quadratic formula and unless I'm doing something wrong I don't com

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I am supposed to find k in the following quadratic equation: (k+3)x^2 + 2kx + 4 = 0. I have tried using the quadratic formula and unless I'm doing something wrong I don't com      Log On


   

Question 109585: I am supposed to find k in the following quadratic equation:
(k+3)x^2 + 2kx + 4 = 0.
I have tried using the quadratic formula and unless I'm doing something wrong I don't come up with k. I have tried using the discriminant to find k with no success. I must be missing something easy.
Thanks for your help, Don
Found 2 solutions by jim_thompson5910, edjones:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Are you trying to find a k that yields one real solution? If you are, then the discriminant must equal zero. So

b%5E2-4ac=0 which means %282k%29%5E2-4%28k%2B3%29%284%29=0


%282k%29%5E2-4%28k%2B3%29%284%29=0

4k%5E2-16k-48=0


Solved by pluggable solver: Quadratic Formula
Let's use the quadratic formula to solve for k:


Starting with the general quadratic


ak%5E2%2Bbk%2Bc=0


the general solution using the quadratic equation is:


k+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29




So lets solve 4%2Ak%5E2-16%2Ak-48=0 ( notice a=4, b=-16, and c=-48)





k+=+%28--16+%2B-+sqrt%28+%28-16%29%5E2-4%2A4%2A-48+%29%29%2F%282%2A4%29 Plug in a=4, b=-16, and c=-48




k+=+%2816+%2B-+sqrt%28+%28-16%29%5E2-4%2A4%2A-48+%29%29%2F%282%2A4%29 Negate -16 to get 16




k+=+%2816+%2B-+sqrt%28+256-4%2A4%2A-48+%29%29%2F%282%2A4%29 Square -16 to get 256 (note: remember when you square -16, you must square the negative as well. This is because %28-16%29%5E2=-16%2A-16=256.)




k+=+%2816+%2B-+sqrt%28+256%2B768+%29%29%2F%282%2A4%29 Multiply -4%2A-48%2A4 to get 768




k+=+%2816+%2B-+sqrt%28+1024+%29%29%2F%282%2A4%29 Combine like terms in the radicand (everything under the square root)




k+=+%2816+%2B-+32%29%2F%282%2A4%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




k+=+%2816+%2B-+32%29%2F8 Multiply 2 and 4 to get 8


So now the expression breaks down into two parts


k+=+%2816+%2B+32%29%2F8 or k+=+%2816+-+32%29%2F8


Lets look at the first part:


x=%2816+%2B+32%29%2F8


k=48%2F8 Add the terms in the numerator

k=6 Divide


So one answer is

k=6




Now lets look at the second part:


x=%2816+-+32%29%2F8


k=-16%2F8 Subtract the terms in the numerator

k=-2 Divide


So another answer is

k=-2


So our solutions are:

k=6 or k=-2





So if k=-2 or k=6, then it will give us one solution for %28k%2B3%29x%5E2+%2B+2kx+%2B+4+

Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
(k+3)x^2 + 2kx + 4 = 0.
kx^2+3x^2+2kx+4=0
kx^2+2kx=-3x^2-4
k(x^2+2x)=-3x^2-4
k=(-3x^2-4)/(x^2+2x)
Ed