Question

Find the value of k for which the equation x2+2(k+1)x+k2=0 has equal roots?

Answer 1

Formula:

Let any given quadratic equation be

ax² + bx + c = 0 ..... (1)

For equal roots, we have

discriminant = 0

i.e., b² - 4ac = 0

Step-by-step explanation:

The given equation is

x² + 2 (k + 1)x + k² = 0

Comparing with equation (1), we get

a = 1, b = 2 (k + 1), c = k²

For equal roots,

b² - 4ac = 0

or, {2 (k + 1)}² - 4 * 1 * k² = 0

or, 4 (k² + 2k + 1) - 4k² = 0

or, 4k² + 4 (2k + 1) - 4k² = 0

or, 4 (2k + 1) = 0

or, 2k + 1 = 0

or, 2k = - 1

or, k = - 1/2

Therefore, the value of k is (- 1/2).

Answer 2

Equal roots Discriminant → b² - 4ac = 0

Then, in x² + 2(k+1)x + x²

a = 1    b = 2(k+1)    c = k²

We substitute these values,

0 = (2(k+1))² - 4(1)(k²)

0 = 4(k+1)² - 4k²

0 = 4(k²+2k+1) - 4k²

0 = 4k² + 8k + 4 - 4k²

(cancelling 4k² and -4k²)

0 = 8k + 4 

8k = -4

k = -4/8

k = -1/2

Hence the value of k = -1/2

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