Find the values of k for roots are real and equal in equation:kx2 + k

1.	Find the values of k for roots are real and equal in equation:kx2 + kx + 1 = – 4x2 – x
Answer» The given equation kx²+ kx + 1 = -4x²– x This can be rewritten as, (k + 4)x² + (k + 1)x + 1 = 0 Now, this in the form of ax²+ bx + c = 0 Where a = (k +4), b = (k + 1), c = 1 For the equation to have real and equal roots, the condition is D = b²– 4ac = 0 ⇒ (k + 1)²– 4(k +4)(1) = 0 ⇒ (k +1)²– 4k – 16 = 0 ⇒ k² + 2k + 1– 4k – 16 = 0 ⇒ k²– 2k – 15 = 0 Now, solving for k by factorization we have ⇒ k²– 5k + 3k – 15 = 0 ⇒ k(k – 5) + 3(k – 5) = 0 ⇒ (k + 3)(k – 5) = 0, k = -3 and k = 5, So, the value of k can either be -3 or 5.

Find the values of k for roots are real and equal in equation:kx2 + kx + 1 = – 4x2 – x

Answer»

The given equation kx²+ kx + 1 = -4x²– x

This can be rewritten as,

(k + 4)x² + (k + 1)x + 1 = 0

Now, this in the form of ax²+ bx + c = 0

Where a = (k +4), b = (k + 1), c = 1

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

D = b²– 4ac = 0

⇒ (k + 1)²– 4(k +4)(1) = 0

⇒ (k +1)²– 4k – 16 = 0

⇒ k² + 2k + 1– 4k – 16 = 0

⇒ k²– 2k – 15 = 0

Now, solving for k by factorization we have

⇒ k²– 5k + 3k – 15 = 0

⇒ k(k – 5) + 3(k – 5) = 0

⇒ (k + 3)(k – 5) = 0,

k = -3 and k = 5,

So, the value of k can either be -3 or 5.