1.

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»

The given equation (3k +1)x+ 2(k +1)x + 1 = 0 is in the form of ax+ bx + c = 0

Where a = (3k +1), b = 2(k + 1), c = 1

For the equation to have real and equal roots, the condition is

D = b– 4ac = 0

⇒ (2(k + 1))– 4(3k +1)(1) = 0

⇒ (k +1)– (3k + 1) = 0 [After dividing by 4 both sides]

⇒ k2 + 2k + 1 – 3k – 1 = 0

⇒ k– k = 0

⇒ k(k – 1) = 0

Either k = 0 Or, k – 3 = 0 ⇒ k = 1,

So, the value of k can either be 0 or 1

Now, using k = 0 in the given quadratic equation we get

(3(0) + 1)x2 + 2(0 + 1)x + 1 = 0

x2 + 2x + 1 = 0

⇒ (x + 1)2 = 0

Thus, x = -1 is the root of the given quadratic equation.

Next, on using k = 1 in the given quadratic equation we get

(3(1) + 1)x2 + 2(1 + 1)x + 1 = 0

4x2 + 4x + 1 = 0

⇒ (2x + 1)2 = 0

Thus, 2x = -1 ⇒ x = -1/2 is the root of the given quadratic equation.



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