Question

If roots of quadratic equation real and equal find k k(2x-5)=x^2-4

Answer 1

Answer:

Step-by-step explanation:

k²(2x -5 ) = x² -4

2xk²- 5k² = x² - 4

x² -2k²x + 5k²-4 = 0

Compare Ax²+Bx + C =0

A = 1 , B = -2k² , C = 5k²-4

And roots are equal then conditions are

B² - 4AC = 0

(-2k²)² - 4×1×(5k²-4) = 0

4k⁴ - 20k²+16 = 0

K⁴ -5k² +4 = 0

k⁴ -4k²-k² +4 = 0

k²(k²-4) - 1 (k²-4) = 0

(k²-4) (k²-1) = 0

Now, k²-4 = 0 or k²-1 = 0

k² = 4 or k² = 1

k = +2 and -2 or k = +1 and -1