# Find the value of of k for which the quadratic equation (k+ 4)x² + (k+ 1)x+1=0 has equal roots and also find the roots

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(k+ 4)x² + (k+ 1)x+1=0

If it has equal roots, b²-4ac = 0

⇒ (k+1)² - 4(k+4)×1 = 0

⇒ k² + 2k + 1 - 4k -16 = 0

⇒ k² - 2k - 15 = 0

⇒ k² - 5k + 3k - 15 = 0

⇒ k(k-5) + 3(k-5) = 0

⇒ (k+3)(k-5) = 0

⇒ k = -3 or 5

if k=-3, the equation is:

x² - 2x + 1 = 0

solve it, you will get (x-1)² = 0 or x=1

if k=5, the equation is:

9x² + 6x + 1 = 0

(3x+1)² = 0

x = -1/3

If it has equal roots, b²-4ac = 0

⇒ (k+1)² - 4(k+4)×1 = 0

⇒ k² + 2k + 1 - 4k -16 = 0

⇒ k² - 2k - 15 = 0

⇒ k² - 5k + 3k - 15 = 0

⇒ k(k-5) + 3(k-5) = 0

⇒ (k+3)(k-5) = 0

⇒ k = -3 or 5

if k=-3, the equation is:

x² - 2x + 1 = 0

solve it, you will get (x-1)² = 0 or x=1

if k=5, the equation is:

9x² + 6x + 1 = 0

(3x+1)² = 0

x = -1/3