find the value of K if equation has equal roots
{k - 12}{x}^{2} -2{k - 12) x + 2
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rajmohammad97:
your answer is wrong
yes he had a wrong sin in the value of b
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{k - 12}{x}^{2} -2{k - 12) x + 2=0
D=0 ( Equal roots )
b^2-4ac=0
{-2(k-12)}^2-4(k-12) (12)=0
{4(k-12)^2} -4(k-12) (12)=0
{4(k-12)} {k-12-12}=0
(k-12) (k-24) =0/4
(k-12) (k-24) =0
k=12, 24
D=0 ( Equal roots )
b^2-4ac=0
{-2(k-12)}^2-4(k-12) (12)=0
{4(k-12)^2} -4(k-12) (12)=0
{4(k-12)} {k-12-12}=0
(k-12) (k-24) =0/4
(k-12) (k-24) =0
k=12, 24
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