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Question
Find the values of k for the following quadratic equation, so that they have two equal roots.
kx (x - 2) + 6 = 0
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Solution
kx(x - 2) + 6 = 0
or kx2 - 2kx + 6 = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = k, b = - 2k and c = 6
Discriminant = b2 - 4ac
= (-2k)2 - 4 (k) (6)
= 4k2 - 24k
k2 - 6k = 0
k (k - 6) = 0
For equal roots,
b2 - 4ac = 0
4k2 - 24k = 0
4k (k - 6) = 0
Either 4k = 0 or k = 6 = 0
k = 0 or k = 6
However, if k = 0, then the equation will not have the terms 'x2' and 'x'.
Therefore, if this equation has two equal roots, k should be 6 only.
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Solution :
Compare x2 + 2x – 9 = 0 with ax2 + bx + c = 0
a = 1, b = 2, c = `square`
∴ b2 – 4ac = (2)2 – 4 × `square` × `square`
Δ = 4 + `square` = 40
∴ b2 – 4ac > 0
∴ The roots of the equation are real and unequal.
