Math, asked by harsha9388, 1 year ago

Find the value of k for which the given equation has real abd equal roots x^2+k(4x+k-1)=0

Answers

Answered by Anonymous
0
★ QUADRATIC RESOLUTION ★

FOR EQUAL ROOTS , D = 0 ,

GIVEN EQUATION → x² + k ( 4x + k - 1 ) = 0
→ x² + 4kx + k² - k = 0
→ x² + x ( 4k ) + ( k² - k ) = 0
D = 0

16k² - 4 ( k² - k ) = 0
16k² - 4k² + 4k = 0
12k² + 4k = 0
k [ 12k +4 ] = 0
∴ k = 0 , k = -1/3

★✩★✩★✩★✩★✩★✩★✩★✩★✩★✩★


Similar questions