Question

# The sum of integral values of p for which the equation √x2−p+2√x2−1=x has real solutions is

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Solution

## √x2−p+2√x2−1=x⋯(1)x2−1≥0⇒x≥1 or x≤−1 And x>0 Therefore, x≥1 Squaring equation (1), we get 4√(x2−p)(x2−1)=p+4−4x2 So, p+4−4x2≥0⇒p≥4(x2−1)⇒p≥0 Squaring again, we get ⇒16(x4−x2−px2+p)=(p+4)2+16x4−8x2(p+4) ⇒8x2(2−p)=p2−8p+16⇒8x2(2−p)=(p−4)2 So, 2−p>0⇒p<2 When p=0, x=1 When p=1, 3√x2−1=x⇒x2−1=x29⇒x=32√2 Sum of integral values of p is 1

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