Question

8: Find the values of k for each of the following quadratic equations, so that they have two equal roots.(i) 2x2 + kx + 3 = 0
(ii) kx (x – 2) + 6 = 0

Solutions:

(i)

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Answer 1

Answer:

Solutions:

(i) 2x2 + kx + 3 = 0

Comparing the given equation with ax2 + bx + c = 0, we get,

a = 2, b = k and c = 3

As we know, Discriminant = b2 – 4ac

= (k)^2 – 4(2) (3)

= k^2 – 24

For equal roots, we know,

Discriminant = 0

k^2 – 24 = 0

k^2 = 24

k = ±√24 = ±2√6

(ii) kx(x – 2) + 6 = 0

or kx2 – 2kx + 6 = 0

Comparing the given equation with ax2 + bx + c = 0, we get

a = k, b = – 2k and c = 6

We know, Discriminant = b2 – 4ac

= ( – 2k)2 – 4 (k) (6)

= 4k2 – 24k

For equal roots, we know,

b2 – 4ac = 0

4k2 – 24k = 0

4k (k – 6) = 0

Either 4k = 0 or k = 6 = 0

k = 0 or k = 6

Answer 2

Answer:

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