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Question

x2+x+1+2k(x2−x−1)=0 is perfect square for how many values of k

A
2
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B
0
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C
1
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D
3
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Solution

The correct option is A 2
Now, x2+x+1+2kx22kx2k=0
(1+2)x2+(12k)x+(12k)=0 ____ (1)
Now, (ax+b)2=a2x2+b2+2abx
Comparing with (1)
2(12k)(1+2k)=(12k)
4(14k2)=1+4k24k
416k2=1+4k24k
20k24k3=0
k=4±16+24040
k=12 or 310
two values

1098420_1141683_ans_d24ae28007ad4ee5976b555f45ae1fd3.jpeg

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