SOLUTION: Hello. I'm having some trouble with the following problem: Find all values of k that ensure that the given equation has exactly one solution. 10x^2 + kx + 8 = 0

Question 196350: Hello.
I'm having some trouble with the following problem:
Find all values of k that ensure that the given equation has exactly one solution.
10x^2 + kx + 8 = 0
Found 2 solutions by solver91311, RAY100:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

In order for

to have exactly one root

must be a perfect square trinomial, so knowing that:

you can clearly see that

where and

so

John

Answer by RAY100(1637) (Show Source): You can put this solution on YOUR website!
10 x^2 +kx +8 =0
.
divide by 10
.
x^2 + (k/10)x + 8/10 =0
.
To get one equal answer, we must have sums squared, (x+a) (x+a) =0,
To complete the square, the coefficient of the 2nd term is halved and then squared
to equal the third term
.
(k/(2*10))^2 = 8/10
k^2/400 = 8/10
k^2 = 320
k= +/- 17.88,,,,(8sqrt5)
.
checking
10 x^2 +(17.88)x +8 =0
x^2 +1.788 x +.8 =0
(x+sqrt.8)(x+sqrt.8) =0
(x +.894) (x+.894) =0
x= -.894, -.894,,, ok
.
checking with k=-17.887,,,,,finds the same,, except x= +.894, +.894 ,,,ok
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