The number of integer solutions of the equation (x2−10)(x2−3x−10)=1 is
Correct Answer: 4
Solution and Explanation
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.
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