1. Find the values of k which the quadratic equation K²x² - 2 (K-D x+4 -0 has real and equal roots. a) K=0 [or] K=! de k=1 or k=1 c) k=-1 or k=1 d) k=-3 or k=1 ) K = k=K 1/ 3 3 3 3
Answers
Answer: Correct option is A)
k
2
x
2
−2(2k−1)x+4=0
⇒ Here, a=k
2
,b=−2(2k−1),c=4
⇒ It is given that roots are real and equal.
∴ b
2
−4ac=0
⇒ [2(2k−1)]
2
−4(k
2
)(4)=0
⇒ 4(4k
2
−4k+1)−16k
2
=0
⇒ 16k
2
−16k+4−16k
2
=0
⇒ −16k=−4
∴ k=
4
1
∴ We can see value of k given in question is correct.
Step-by-step explanation:
k2x2 – 2 (k – 1)x + 4 = 0
Sol. k2x2 – 2 (k – 1)x + 4 = 0
Comparing with ax2 + bx + c = 0 we have a = k2, b = – 2 (k – 1), c = 4
We know that,
∆ = b2 – 4ac
= [– 2 (k – 1)]2 – 4 (k2) (4)
= (– 2k + 2)2 – 16k2
= 4k2 – 8k + 4 – 16k2
= – 12k2 – 8k + 4
∵ The roots of given equation are real and equal.
∴ ∆ must be zero.
∴ – 12k2 – 8k + 4 = 0
∴ – 4 (3k2 + 2k – 1) = 0
∴ 3k2 + 3k – k – 1 = -0/4
∴ 3k (k + 1) – 1 (k + 1) = 0
∴ (k + 1) (3k – 1) = 0
∴ k + 1 = 0 or 3k – 1 = 0
∴ k = – 1 or 3k = 1
∴ k = -1 or k = 1/3