Question

For what value of k will the following quadratic equation give real roots? Also, find the solution for that value of k-
(k-12)x²+2(k-12)x+2=0
Ans.=14; solution=-1 as given in textbook.
Please solve it quickly! I have to give it as homework tomorrow.

Answer 1

For real root
b²-4ac ≥0
(k-12)x²+2(k-12)x+2=0
I this equation
a=(k-12)
but b=2(k-12) and c= 2

now

(2(k-12))² - 4× (k-12)×2 ≥0
4(k²+144-24k) -(8k-96) ≥0
4k²+576-96k -8k +96≥0
4k²-104k +672≥0
k²-26k+168≥0
k²-14k-12k+168≥0
k(k-14)-12(k-14)≥0
(k-12)(k-14)≥0

either
k≥12 or
k≥14