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Question

For what integer value/s, of k, the quadratic equation (1k)x24kx+2=0 has real and equal roots?


A
1
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B
12,1
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C
12
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D
None of these
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Solution

The correct option is A 1
(1k)x24kx+2=0

a=1k,b=4k,c=2

Since the quadratic equation has real and equal roots,

D=b24ac=0

(4k)24×(1k)×2=0

16k28(1k)=0

16k28+8k=0

2k2+k1=0

2k2+2kk1=0

(k+1)(2k1)=0

k=1,12

Since, we have to find the inetger value

Hence, k=-1.

(a) is correct

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