Asked by Euler
Prove that the roots of the equation kx^2+(2k+4)x+8=0 are real for all values of k.
Answers
Answered by
Reiny
For a quadratic of the form ax^2 + bx + c = 0 to have real roots,
the discriminant b^2 - 4ac ≥ 0
(2k+4)^2 - 4k(8) ≥ 0
4k^2 + 16k + 16 - 32k ≥ 0
k^2 - 4k + 4 ≥ 0
(k-2)^2 ≥ 0
(k - 2) ≥ 0
k-2 ≥ 0 OR -k+2 ≥ 0
k ≥ 2 OR -k ≥ -2 ----> k ≤ 2
Well, k≥ 2 OR k≤2 would refer to all values of k
the discriminant b^2 - 4ac ≥ 0
(2k+4)^2 - 4k(8) ≥ 0
4k^2 + 16k + 16 - 32k ≥ 0
k^2 - 4k + 4 ≥ 0
(k-2)^2 ≥ 0
(k - 2) ≥ 0
k-2 ≥ 0 OR -k+2 ≥ 0
k ≥ 2 OR -k ≥ -2 ----> k ≤ 2
Well, k≥ 2 OR k≤2 would refer to all values of k
Answered by
oobleck
Recall the discriminant, b^2-4ac
If it is positive, there are two real roots.
So, is (2k+4)^2 - 4*k*8 always positive?
If it is positive, there are two real roots.
So, is (2k+4)^2 - 4*k*8 always positive?
Answered by
Anonymous
as (k-2) is squared its always positive...so the discriminant is always greater than zero.???
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.