SOLUTION: Find the number k such that the given equation has exactly one real solution: kx^2+28x+4=0

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Question 590370: Find the number k such that the given equation has exactly one real solution:
kx^2+28x+4=0

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
For kx%5E2%2B28x%2B4+=+0 to have exactly one real solution. the factors of this equation have to be equal.
For example %28ax%2Bb%29%28ax%2Bb%29 so that b+=+2 and a+=+7 to get:
7x%2B2%29%287x%2B2%29+=+49x%5E2%2B28x%2B4 and the solution(s) is(are) x+=+-0.28571434
So k+=+49
But you could do it this way:
kx%5E2%2B28x%2B4+=+0 Complete the square. First, divide by k.
%28kx%5E2%2B28x%2B4%29%2F%28k%29%29=0 Simplify.
x%5E2%2B28x%2Fk+%2B4%2Fk+=+0 To complete the square, the constant term must equal the square of half the x-coefficient, so %28%281%2F2%29%2828%2Fk%29%29%5E2+=+%2814%2Fk%29%5E2=196%2Fk%5E2
Set this equal to the constant term 4%2Fk and solve for k.
196%2Fk%5E2+=+4%2Fk Simplify.
196%2Fk+=+4
k+=+196%2F4
k+=+49
So the equation is:
49x%5E2%2B28x%2B4+=+0
graph%28400%2C400%2C-2%2C2%2C-5%2C5%2C49x%5E2%2B28x%2B4%29