Question

If - 4 is a root of the equation x^2 + 2x + 4p = 0, find the value of k for which the quadratic equation x^2 + px( 1 + 3k ) + 7( 3 + 2k ) = 0 has equal roots.

Answer 1

hope this will help u.. :-)

Answer 2

Given Quadratic Equation is x^2 + 2x + 4p = 0.

(i)

Given that -4 is a root of the equation.

Substitute x = -4, we get

⇒ (-4)^2 + 2(-4) + 4p = 0

⇒ 16 - 8 + 4p = 0

⇒ 8 = -4p

⇒ p = -2.

(ii)

Given that x^2 + px(1 + 3k) + 7(3 + 2k) = 0 has equal roots.

Substitute p = -2, then equation becomes.

⇒ x^2 - 2x(1 + 3k) + 7(3 + 2k) = 0

⇒ x^2 - x(2 - 6k) + 7(3 + 2k) = 0

Here, a = 1, b = -2 + 6k, c = 7(3 + 2k) = 21 + 14k.

Given Equation has equal roots.

∴ Δ = 0

⇒ b² - 4ac = 0

⇒ (-2 + 6k)² - 4(1)(21 + 14k) = 0

⇒ 4 + 36k² + 24k - 84 - 56k = 0

⇒ 36k² - 32k - 80 = 0

⇒ 9k² - 8k - 20 = 0

⇒ 9k² - 18k + 10k - 20 = 0

⇒ 9k(k - 2) + 10(k - 2) = 0

⇒ (9k + 10)(k - 2) = 0

⇒ k = -10/9,2

Therefore, the value of k = -10/9, 2.

Hope it helps!