Question

The number of integer solutions of the equation (x2−10)(x2−3x−10)=1 is

Answer 1

Correct Answer: 4

Case 1: When \(x^2 - 3x - 10 = 0\) and \(x^2 - 10 \ne 0\)

Solve the quadratic: \[ x^2 - 3x - 10 = 0 \] Factorizing: \[ (x - 5)(x + 2) = 0 \] So, \(x = 5\) or \(x = -2\)

Now check that neither value satisfies \(x^2 - 10 = 0\), which is true, so both values are valid.

Case 2: When \(x^2 - 10 = 1\)

Then: \[ x^2 = 11 \] This has no integer solutions. So no values of \(x\) from this case.

Case 3: When \(x^2 - 10 = -1\) and \(x^2 - 3x - 10\) is even

Then: \[ x^2 = 9 \] So: \[ (x + 3)(x - 3) = 0 \Rightarrow x = -3 \text{ or } x = 3 \]

Now check if \(x^2 - 3x - 10\) is even for both values:

• For \(x = -3\): \(x^2 - 3x - 10 = 9 + 9 - 10 = 8\), which is even ✅
• For \(x = 3\): \(x^2 - 3x - 10 = 9 - 9 - 10 = -10\), which is even ✅

So both values are valid.

Conclusion: The valid integer solutions for \(x\) are: \[ x = -3, -2, 3, 5 \] That is, a total of 4 values satisfy the conditions.