Question

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

Answer 1

The correct answer is: 4
For the given equation, we have: 
\((x^2 - 10)^{(x^2 - 3x - 10)} = 1\) 
Since any non-zero number raised to the power of 0 is 1,we can immediately see that one possible solution is when the base \((x^2 - 10)\) is equal to 1: 
\(x^2 - 10 = 1\) 
\(x^2=11\) 
This gives us two solutions for \(x: x = \sqrt{11}\space{and}\space x=-\sqrt{11}\). 
Now,let's consider the case where the exponent \((x^2 - 3x - 10)\) is equal to 0: 
\(x^2 - 3x - 10 = 0\) 
This is a quadratic equation that can be factored: 
(x-5)(x+2)=0 
This gives us two more solutions: x=5 and x=-2. 
However,we need to check whether these solutions satisfy the original equation: 
For \(x=\sqrt{11}\space{and}\space x=-\sqrt{11}\): 
\((x^2-10)^{(x^2-3x-10)}=(11-10)^0=1^0=1\) 
For x = 5 and x = -2: 
\((x^2-10)^{(x^2-3x-10)}=(25-10)^{(25-3(5)-10)}=15^0=1 \)
All four solutions satisfy the given equation, so the total number of integer solutions is 4: \(x = \sqrt{11}, x = -\sqrt{11}, x = 5\), and x = -2. 

Answer 2

When \(x^2-3x-10=0 \space and \space x^2-10≠0\)
\(x^2-3x-10=0 \space or, \space (x-5)(x+2)=0\)
or, x= 5 or -2
\(x^2-10 = 1\)
\(x^2-11 = 0\)
No integer solutions
\(x^2 - 10=-1\) and \(x^2-3x-10\) is even -
\(x^2-9=0\)
or, \((x+3)(x-3)=0\)
or, \(x= -3 \space and \space 3\)
for \(x= -3\) and \(+3 x^2 - 3x -10\) is even
In total 4 values of x satisfy the equations