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