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Ankush Raj from S.V.D Public School, asked a question
Subject: Math , asked on 27/10/12

FIND THE VALUE OF K FOR WHICH THE ROOTS OF THE EQUATION (K+4) x2 + (K+1) x + 1 = 0 are real and equal.

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Komal Singh From Delhi Public School, added an answer, on 29/10/12
633 helpful votes in Math

since for the real and equal roots ,

b2 -4ac =0

(k+1)2 - 4 x (K+4) x 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-5) (k+3) =0

then , k = 5   and k =-3

HOPE THIS HELPS...

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Param Sukhadia From Kendriya Vidhyalaya No. 1, added an answer, on 27/10/12
645 helpful votes in Math

 (k+i)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-5)(k+3)=0

k=5 and k=-3

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Param Sukhadia From Kendriya Vidhyalaya No. 1, added an answer, on 27/10/12
645 helpful votes in Math

[k+1]x - 2[k-1]x+1=0

Now, for a given equation to have real and equal roots, its discriminant, D must be equal to 0.

⇒ D = b 2 - 4ac = 0

⇒ [-2(k-1)] 2 - 4(k + 1)1 = 0

⇒ 4(k-1) 2 = 4(k + 1)

⇒ 4k + 4 - 8k = 4k + 4

⇒ 4k -12k = 0

⇒ 4k (k - 3) = 0

⇒ k = 0, 3

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