Find the value of k for (k+1)x 2 -2(k-1)x+1=0,so that it has two equal roots
Answers
Answered by
1
(k+1)x²-2(k-1)x+1=0 --------(1)
Let, α,β are the roots of (1). Then, α=β.
∴,α+β=-[-2(k-1)/(k+1)]
or, 2α=2(k-1)/(k+1)
or, α=k-1/k+1
or, α²=(k-1)²/(k+1)² --------------(2)
α.β=1/k+1
or, α.α=1/k+1
or, α²=1/k+1 ---------------------(3)
From (2) and (3)-
(k-1)²/(k+1)²=1/(k+1)
or, (k-1)²/(k+1)=1
or, (k-1)²=k+1
or, k²-2k+1=k+1
or, k²-2k-k+1-1=0
or, k²-3k=0
or, k(k-3)=0
either, k=0
or, k-3=0
or, k=3
∴, k=0,3
Let, α,β are the roots of (1). Then, α=β.
∴,α+β=-[-2(k-1)/(k+1)]
or, 2α=2(k-1)/(k+1)
or, α=k-1/k+1
or, α²=(k-1)²/(k+1)² --------------(2)
α.β=1/k+1
or, α.α=1/k+1
or, α²=1/k+1 ---------------------(3)
From (2) and (3)-
(k-1)²/(k+1)²=1/(k+1)
or, (k-1)²/(k+1)=1
or, (k-1)²=k+1
or, k²-2k+1=k+1
or, k²-2k-k+1-1=0
or, k²-3k=0
or, k(k-3)=0
either, k=0
or, k-3=0
or, k=3
∴, k=0,3
Similar questions