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Find value of k, (k+1)X2+2(k-1)x+(k-2) |
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Answer» The given equation is (k+1)x 2 −2(k−1)x+1=0 comparing it with ax 2 +bx+c=0 we geta=(k+1),b=−2(k−1) and c=1∴ Discriminant,D=b 2 −4ac=4(k−1) 2 −4(k+1)×1=4(k 2 −2k+1)−4k−4⇒4k 2 −8k+4−4k−4=4k 2 −12kSince roots are real and equal, soD=0⇒4k 2 −12k=0⇒4k(k−3)=0⇒ either k=0 or k−3=0⇒4k(k−3)=0Hence, k=0,3. k=2 |
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