Find the value of k if the equation (k-12)x square +2(k+2)x+2=0
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2
We know that
b2-4ac=0 (given)
(2k-24)2- 4(k-12)(2)=0
4k2-96k+576-8k+96=0
4k2-104k+672/4=0
K2-26k+168=0
(sum=-26,product=168)
K2-12k-14k+168=0
K(k-12)-14(k-12)=0
k-12=0 (or) k-14=0
k=12(or)k=14
b2-4ac=0 (given)
(2k-24)2- 4(k-12)(2)=0
4k2-96k+576-8k+96=0
4k2-104k+672/4=0
K2-26k+168=0
(sum=-26,product=168)
K2-12k-14k+168=0
K(k-12)-14(k-12)=0
k-12=0 (or) k-14=0
k=12(or)k=14
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8
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