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Question

The sum of integral values of p for which the equation x2p+2x21=x has real solutions is

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Solution

x2p+2x21=x(1)x210x1 or x1
And x>0
Therefore, x1
Squaring equation (1), we get
4(x2p)(x21)=p+44x2
So,
p+44x20p4(x21)p0
Squaring again, we get
16(x4x2px2+p)=(p+4)2+16x48x2(p+4)
8x2(2p)=p28p+168x2(2p)=(p4)2
So,
2p>0p<2
When p=0,
x=1
When p=1,
3x21=xx21=x29x=322
Sum of integral values of p is 1

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