find the value of k for which given equation has real and equal roots 1.(K-12)x square +(K-12)x+2=0
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The equation has equal roots
therrfore b2 - 4ac =0
{ 2(k-12) }2 - 4 ( k-12) 2 =0
4 ( k2 - 24k + 144 ) - 8k + 96 =0
4k2 - 96k + 576 - 8k + 96 = 0
4k2 - 104 k +672 = 0
k2 -26k +168 =0
k 2 -14k-12k +168 = 0
k ( k-14) -12 (k-14) =0
(k-12)(k-14)=0
k=12 or k=14
therrfore b2 - 4ac =0
{ 2(k-12) }2 - 4 ( k-12) 2 =0
4 ( k2 - 24k + 144 ) - 8k + 96 =0
4k2 - 96k + 576 - 8k + 96 = 0
4k2 - 104 k +672 = 0
k2 -26k +168 =0
k 2 -14k-12k +168 = 0
k ( k-14) -12 (k-14) =0
(k-12)(k-14)=0
k=12 or k=14
angel988390:
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some question
K square x square- 2 (k-1)x+4
k square x square -2 (k-1)x+4=0
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