Question

Find the value of k for which equation has real and equal roots:
(K+1) x2+2(k+3)x + (k+8)=0

Answer 1

Step-by-step explanation:

The given equation (k +1)x2 + 2(k +3)x + (k +8) = 0 is in the form of ax2 + bx + c = 0

Where a = (k +1), b = 2(k + 3), c = (k + 8)

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

D = b2 – 4ac = 0

⇒ (2(k + 3))2 – 4(k +1)(k + 8) = 0

⇒ 4(k +3)2 – 4(k2 + 9k + 8) = 0

⇒ (k + 3)2 – (k2 + 9k + 8) = 0

⇒ k2 +6k + 9 – k2 – 9k – 8 = 0

⇒ -3k + 1 = 0

⇒ k = 1/3

So, the value of k is 1/3.

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