Question

The value of k for which the Q.E (k+4).r? +(k+1)x+1=0 has equal roots
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Answer 1

Answer:

this is the answer in this question

Answer 2

Given quadratic equation:

(k+4)x²+(k+1)x+1=0.

Since the given quadratic equation has equal roots, its discriminant should be zero.

D = 0

→ (k+1)²-4 x (k+4) × 1=0

⇒k²+2k+1-4k-16=0

⇒k²-2k-15-0

⇒k²-5k+3k-15=0

⇒(k-5) (k+3)=0

⇒k-5-0 or k+3=0

⇒k=5 or -3

Thus, the values of k are 5 and -3.

For k = 5:

(k+4)x²+(k+1) x+1=0

⇒9x²+6x+1=0

⇒(3x)²+2(3x)+1=0

⇒(3x+1)²=0

⇒x=-1/3, -1/3

For k = -3:

(k+4)x²+(k+1)x+1=0

⇒x²-2x+1=0

⇒(x-1)²=0

⇒x=1,1

Thus, the equal root of the given quadratic equation is either 1 or -13.

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