Question

The number of integer solutions of the equation \((x^2−10)^{(x2−3x−10)}=1\) is

Answer 1

We know that any number raised to the power 0 is 1 (as long as the base ≠ 0), and 1 raised to any power is also 1.

So we consider the following cases that make the equation true:

• \( (x^2 - 10) = 1 \)
• \( (x^2 - 10) = -1 \) (may need further checking)
• \( (x^2 - 3x - 10) = 0 \)

 Case 1: Base = 1

If \( x^2 - 10 = 1 \), then: \[ x^2 = 11 \Rightarrow x = \pm \sqrt{11} \] These are two real (irrational) solutions.

Check the original equation: \[ (\sqrt{11}^2 - 10)^{(\sqrt{11}^2 - 3\sqrt{11} - 10)} = (1)^{\text{some value}} = 1 \] So both \( x = \sqrt{11} \) and \( x = -\sqrt{11} \) are valid.

 Case 2: Exponent = 0

Set exponent equal to zero: \[ x^2 - 3x - 10 = 0 \Rightarrow (x - 5)(x + 2) = 0 \Rightarrow x = 5,\ x = -2 \] Now verify these in the original equation:

• For \( x = 5 \): \( (25 - 10)^0 = 15^0 = 1 \) ✔
• For \( x = -2 \): \( (4 - 10)^0 = (-6)^0 = 1 \) ✔

 

 Summary of Valid Solutions

• \( x = \sqrt{11} \)
• \( x = -\sqrt{11} \)
• \( x = 5 \)
• \( x = -2 \)

All four values satisfy the given equation: \[ \boxed{(x^2 - 10)^{(x^2 - 3x - 10)} = 1} \]

 Final Answer

Total number of solutions: \[ \boxed{4} \]

Answer 2

Given:
\(x^2 - 3x - 10 = 0\) and \(x^2 - 10 \ne 0\)

Step 1: Solve the quadratic equation
\(x^2 - 3x - 10 = 0\)
Factorizing:
\(x^2 - 3x - 10 = (x - 5)(x + 2) = 0\)
Therefore, \(x = 5\) or \(x = -2\)

Step 2: Check the condition \(x^2 - 10 \ne 0\)
For \(x = 5\):
\(x^2 - 10 = 25 - 10 = 15 \ne 0\) ✅
For \(x = -2\):
\(x^2 - 10 = 4 - 10 = -6 \ne 0\) ✅
So both values \(x = 5\) and \(x = -2\) satisfy the conditions.

Step 3: Solve \(x^2 - 10 = 1\)
\(x^2 = 11\)
\(x = \pm\sqrt{11}\)
These are not integers ❌

Step 4: Solve \(x^2 - 10 = -1\)
\(x^2 = 9\)
\(x = \pm3\)

Check if \(x^2 - 3x - 10\) is even for these values:

For \(x = 3\):
\(x^2 - 3x - 10 = 9 - 9 - 10 = -10\) (Even) ✅ 

For \(x = -3\):
\(x^2 - 3x - 10 = 9 + 9 - 10 = 8\) (Even) ✅

Final Answer:
The values of \(x\) that satisfy the given conditions are:
\(x = -3, -2, 3, 5\)

Total number of values = 4