Question

if 2 x^2-(2 + K)X + K=0 find the value of k for which the roots of the equation are real and equal

Answer 1

D = b^2 -4ac
0>(2+k)^2-4 ×k×2
0>4 + k^2+ 4k-8k
0>k^2-4k+4
0>(k-2)^2
0>k-2
k=2

Answer 2

Hi Mate!!!

For real and equal roots Descrimnant ( D) must be equal to zero

D = √ {( 2 + k )² - 4 ( 2 ) ( k )}

D = 0 for real and equal roots.

√{ ( 2 + k )² - 8 k } = 0

squaring both sides

4 + k² + 4k - 8k = 0

k² - 4k + 4 = 0

k = 4 + √ ( 16 - 16 ) / 2 or k = 4 - √ ( 16 - 16 ) / 2

k = 2 or k = 2

Have a nice time...