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Question
Find the values of k for which the roots are real and equal in each of the following equation:
k2x2 - 2(2k - 1)x + 4 = 0
Solution
The given equation is k2x2 - 2(2k - 1)x + 4 = 0
The given equation is in the form of ax2 + bx + c = 0
where a = k2, b = -2(2k - 1) and c = 4
therefore, the discriminant
D = b2 - 4ac
= (-2(2k - 1))2 - 4 x (k2) x (4)
= 4(2k - 1)2 - 16k2
= 4(4k2 + 1 - 4k) - 16k2
= 16k2 + 4 - 16k - 16k2
= 4 - 16k
∵ Roots of the given equation are real and equal
∴ D = 0
⇒ 4 - 16k = 0
⇒ -16k = -4
`rARrk=(-4)/-16`
⇒ k = 1/4
Hence, the value of K = 1/4
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