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

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)x²+ 2(k +1)x + k = 0 is in the form of ax²+ 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 = b²– 4ac = 0 ⇒ (2(k + 1))²– 4(3k +1)(k) = 0 ⇒ 4(k +1)²– 4(3k² + k) = 0 ⇒ (k + 1)²– k(3k + 1) = 0 ⇒ 2k²– k – 1 = 0 Now, solving for k by factorization we have ⇒ 2k²– 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\(\frac{-1}{2}\).

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)x²+ 2(k +1)x + k = 0 is in the form of ax²+ 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 = b²– 4ac = 0

⇒ (2(k + 1))²– 4(3k +1)(k) = 0

⇒ 4(k +1)²– 4(3k² + k) = 0

⇒ (k + 1)²– k(3k + 1) = 0

⇒ 2k²– k – 1 = 0

Now, solving for k by factorization we have

⇒ 2k²– 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\(\frac{-1}{2}\).