1.

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

Answer»

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

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

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

D = b– 4ac = 0

⇒ (2(k + 1))– 4(3k +1)(k) = 0

⇒ 4(k +1)– 4(3k2 + k) = 0

⇒ (k + 1)– k(3k + 1) = 0

⇒ 2k– k – 1 = 0

Now, solving for k by factorization we have

⇒ 2k– 2k + k – 1 = 0

⇒ 2k(k – 1) + 1(k – 1) = 0

⇒ (k – 1)(2k + 1) = 0,

k = 1 and k = \(\frac{-1}{2}\),

So, the value of k can either be 1 or\(\frac{-1}{2}\).



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