Find the value of k if the equation x^2-2 (k+1)x+kx=0 has equal roots
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For equal roots
D=0
b^2-4ac=0
{-2(k+1)}^2-4k=0
(-2k-2)^2-4k=0
4k^2+8k+4=0
4k^2+4k+4k+4=0
4k(k+1)+4(k+1)=0
(k+1)(4k+4)=0
either k=-1 or k=-4/4=-1
answer
D=0
b^2-4ac=0
{-2(k+1)}^2-4k=0
(-2k-2)^2-4k=0
4k^2+8k+4=0
4k^2+4k+4k+4=0
4k(k+1)+4(k+1)=0
(k+1)(4k+4)=0
either k=-1 or k=-4/4=-1
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