1.

Find the values of k for roots are real and equal in equation:(k + 1)x2 – 2(3k + 1)x + 8k + 1 = 0

Answer»

The given equation (k +1)x– 2(3k +1)x + 8k + 1 = 0 is in the form of ax+ bx + c = 0

Where a = (k +1), b = -2(3k + 1), c = 8k + 1

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

D = b– 4ac = 0

⇒ (-2(3k + 1))– 4(k +1)(8k + 1) = 0

⇒ 4(3k +1)– 4(k + 1)(8k + 1) = 0

⇒ (3k + 1)– (k + 1)(8k + 1) = 0

⇒ 9k+ 6k + 1 – (8k+ 9k + 1) = 0

⇒ 9k+ 6k + 1 – 8k– 9k – 1 = 0

⇒ k– 3k = 0

⇒ k(k – 3) = 0

Either k = 0 Or, k – 3 = 0 ⇒ k = 3,

So, the value of k can either be 0 or 3



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