Question

Find the exact values of k

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Answer 1

Question :-

Find the exact value of k for which the equation 2kx + (k+1)x + 1 = 0 has equal roots.

Solution :-

For an equation to have equal roots,

b^2 - 4ac = 0

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From the equation,

• b^2 = (k+1)^2
• a = 2k
• c = 1

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Now,

D = (k+1)^2 - 4(2k)(1)

= k^2 + 2k + 1 - 8k = 0

= k^2 - 6k + 1 = 0

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Solving by quadratic formula,

k = 3 + 2√2 or

k = 3 - 2√2

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Question :-

Find the range of value for k for which the equation 2x^2 + 6x + k has no real roots.

Solution

For an equation to have no real roots

b^2 - 4ac < 0

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From the equation,

• b^2 = 6^2
• a = 2
• c = k

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So,

b^2 - 4ac = 36 - 4(2)(k)

= 36 - 8k

= -8k < -36

= k > 36/8

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