Question

Q16. Find the values of k for which the quadratic equation
(3k + 1)x² + 2(k + 1)x + 1 = 0 has equal roots. Also find the roots.​

Answer 1

Step-by-step explanation:

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Answer 2

Given:-

• The quadratic equation (3k + 1)x² + 2(K + 1)x + 1 = 0 has equal roots.

To find:-

• Find the value of k.?

Solutions:-

• The given quadratic equation (3k + 1)x² + 2(K + 1)x + 1 = 0 and root are real and equal.

Here,

• a = 3k + 1
• b = 2(k + 1)
• c = 1

We know that:

• D => b² - 4ac

Putting the value of a = 3k + 1, b = 2(k + 1) and c = 1

• D => b² - 4ac

=> [2(k + 1)]² - 4(3k + 1) (1)

=> 4(k² + 2k + 1) - 12k - 4

=> 4k² + 8k + 4 - 12k - 4

=> 4k² - 4k

The given equation will have real and equal roots,

• D = 0

Thus,

=> 4k² - 4k = 0

=> 4k(k - 1) = 0

=> k = 0 Or k - 1 = 0

=> k = 0 Or k = 1

Therefore,

The value of k is o or 1.

Now,

• for k = 0, the equation.

=> x² + 2x + 1 = 0

=> x² + x + x + 1 = 0

=> x(x + 1) + 1(x + 1) = 0

=> x = -1 , -1

• for k = 1, the equation.

=> 4x² + 4x + 1 = 0

=> 4x² + 2x + 2x + 1 = 0

=> 2x(2x + 1) + 1(2x + 1) = 0

=> (2x + 1)² = 0

=> x = -1/2 , -1/2

Hence, the roots of the equation are -1 and -1/2