SOLUTION: Given that the equation {{{kx^2+12x+k=0}}} where k is a positive constant and has equal roots. Find the value of k.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Given that the equation {{{kx^2+12x+k=0}}} where k is a positive constant and has equal roots. Find the value of k.       Log On


   



Question 664293: Given that the equation kx%5E2%2B12x%2Bk=0 where k is a positive constant and has equal roots. Find the value of k.
Found 2 solutions by lynnlo, vleith:
Answer by lynnlo(4176) About Me  (Show Source):
You can put this solution on YOUR website!
k=-12x/x^2+1

Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
You are told to find k such that the given equation results in two equal roots.
That says the resulting answer will end up being in the form %28ax%2Bb%29%5E2+=+0
So Let's expand that
%28ax%2Bb%29%5E2+=+0
a%5E2x%5E2+%2B+2abx+%2B+b%5E2+=+0
Now lets match that up with the given problem
We see that k+=+a%5E2, 12+=+2ab, and k+=+b%5E2
We can see from this that 6=ab and also that a%5E2+=+b%5E2
Using the a%5E2+=+b%5E2, we can see that either a=b or a=-b
Since a*b is positive (6), then a must be equal to b.
If a=b and ab =6, then a and b are both sqrt%286%29
So what does than make k? k is a^2 = 6.
Check our answer
kx%5E2%2B12x%2Bk=0
6x%5E2+%2B+12x+%2B+6+=+0
x%5E2+%2B2x+%2B1+=+0
%28x%2B1%29%5E2+=+0