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Question

For what integer value/s of k, the quadratic equation (x1)(xk)=4 has real and equal roots. (k > 0)

A
5, -3
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B
5
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C
5, 2
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D
2, -3
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Solution

The correct option is B 5
(x1)(xk)=4

Converting into standard from

x2kxx+k=4

x2(k+1)x+(k+4)=0

a=1,b=(k+1),c=k+4

Since the quadratic equation has real and equal roots,

D=b24ac=0

((k+1))24×1×(k+4)=0

k2+2k+14K16=0

k22k15=0

k25k+3k15=0

(k5)(k+3)=0

k=5,3

Since, we have to find the positive value of k.

Hence, k =5

(b) is correct


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