Question:
If x is an integer, how many possible values of x satisfy the equation: \((x - 2)^{2(x + 1)} = 1\)?
If x is an integer, how many possible values of x satisfy the equation: \((x - 2)^{2(x + 1)} = 1\)?
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When solving Diophantine equations (equations where you seek integer solutions), always look for constraints. Here, the factor \((x-2)\^{}2\) being a perfect square is a strong constraint. It must be non-negative, which immediately eliminates one of the two possible factor pairs for 1.
Updated On: Oct 3, 2025
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
We are looking for integer solutions to a polynomial equation. Since the variables are restricted to integers, we can use the properties of integer factorization to solve the equation. The product of two integers equals 1 only under specific conditions.
Step 2: Key Formula or Approach:
The given equation is \((x-2)\^{}2(x+1)=1\).
Since \(x\) is an integer, \(x-2\) and \(x+1\) are also integers.
Therefore, \((x-2)\^{}2\) is an integer. Let \(A = (x-2)\^{}2\) and \(B = (x+1)\).
The equation is \(A \cdot B = 1\).
For the product of two integers \(A\) and \(B\) to be 1, there are two possibilities:
1. \(A=1\) and \(B=1\).
2. \(A=-1\) and \(B=-1\).
Step 3: Detailed Explanation:
Let's analyze the two possible cases.
Case 1: \(A=1\) and \(B=1\)
This gives us a system of two equations:
1) \((x-2)\^{}2 = 1\)
2) \(x+1 = 1\)
From the second equation, \(x+1 = 1\), we find \(x = 0\).
Now, we must check if this value of \(x\) also satisfies the first equation.
Substitute \(x=0\) into \((x-2)\^{}2 = 1\):
\[ (0-2)\^{}2 = (-2)\^{}2 = 4 \] Since \(4 \neq 1\), the value \(x=0\) is not a solution to the system. Therefore, this case yields no integer solutions.
Case 2: \(A=-1\) and \(B=-1\)
This gives us the system:
1) \((x-2)\^{}2 = -1\)
2) \(x+1 = -1\)
Let's look at the first equation, \((x-2)\^{}2 = -1\).
Since \(x\) is an integer, \(x-2\) is also an integer. The square of any real number (including integers) must be non-negative (\(\ge 0\)). It is impossible for the square of an integer to be -1.
Therefore, this case yields no solutions.
Since neither case provides a valid integer solution for \(x\), there are no integer values of \(x\) that satisfy the given equation.
Alternative Method (Polynomial Roots):
Expand the equation:
\[ (x\^{}2 - 4x + 4)(x+1) = 1 \] \[ x\^{}3 - 4x\^{}2 + 4x + x\^{}2 - 4x + 4 = 1 \] \[ x\^{}3 - 3x\^{}2 + 3 = 0 \] By the Rational Root Theorem, if there is an integer root for this polynomial, it must be a divisor of the constant term, which is 3. The divisors of 3 are \(\pm 1, \pm 3\). Let's test these values:
- For \(x=1\): \(1\^{}3 - 3(1)\^{}2 + 3 = 1 - 3 + 3 = 1 \neq 0\).
- For \(x=-1\): \((-1)\^{}3 - 3(-1)\^{}2 + 3 = -1 - 3 + 3 = -1 \neq 0\).
- For \(x=3\): \(3\^{}3 - 3(3)\^{}2 + 3 = 27 - 27 + 3 = 3 \neq 0\).
- For \(x=-3\): \((-3)\^{}3 - 3(-3)\^{}2 + 3 = -27 - 27 + 3 = -51 \neq 0\).
None of the possible integer roots satisfy the equation. This confirms that there are no integer solutions.
Step 4: Final Answer:
There are 0 possible integer values of x that satisfy the equation.
We are looking for integer solutions to a polynomial equation. Since the variables are restricted to integers, we can use the properties of integer factorization to solve the equation. The product of two integers equals 1 only under specific conditions.
Step 2: Key Formula or Approach:
The given equation is \((x-2)\^{}2(x+1)=1\).
Since \(x\) is an integer, \(x-2\) and \(x+1\) are also integers.
Therefore, \((x-2)\^{}2\) is an integer. Let \(A = (x-2)\^{}2\) and \(B = (x+1)\).
The equation is \(A \cdot B = 1\).
For the product of two integers \(A\) and \(B\) to be 1, there are two possibilities:
1. \(A=1\) and \(B=1\).
2. \(A=-1\) and \(B=-1\).
Step 3: Detailed Explanation:
Let's analyze the two possible cases.
Case 1: \(A=1\) and \(B=1\)
This gives us a system of two equations:
1) \((x-2)\^{}2 = 1\)
2) \(x+1 = 1\)
From the second equation, \(x+1 = 1\), we find \(x = 0\).
Now, we must check if this value of \(x\) also satisfies the first equation.
Substitute \(x=0\) into \((x-2)\^{}2 = 1\):
\[ (0-2)\^{}2 = (-2)\^{}2 = 4 \] Since \(4 \neq 1\), the value \(x=0\) is not a solution to the system. Therefore, this case yields no integer solutions.
Case 2: \(A=-1\) and \(B=-1\)
This gives us the system:
1) \((x-2)\^{}2 = -1\)
2) \(x+1 = -1\)
Let's look at the first equation, \((x-2)\^{}2 = -1\).
Since \(x\) is an integer, \(x-2\) is also an integer. The square of any real number (including integers) must be non-negative (\(\ge 0\)). It is impossible for the square of an integer to be -1.
Therefore, this case yields no solutions.
Since neither case provides a valid integer solution for \(x\), there are no integer values of \(x\) that satisfy the given equation.
Alternative Method (Polynomial Roots):
Expand the equation:
\[ (x\^{}2 - 4x + 4)(x+1) = 1 \] \[ x\^{}3 - 4x\^{}2 + 4x + x\^{}2 - 4x + 4 = 1 \] \[ x\^{}3 - 3x\^{}2 + 3 = 0 \] By the Rational Root Theorem, if there is an integer root for this polynomial, it must be a divisor of the constant term, which is 3. The divisors of 3 are \(\pm 1, \pm 3\). Let's test these values:
- For \(x=1\): \(1\^{}3 - 3(1)\^{}2 + 3 = 1 - 3 + 3 = 1 \neq 0\).
- For \(x=-1\): \((-1)\^{}3 - 3(-1)\^{}2 + 3 = -1 - 3 + 3 = -1 \neq 0\).
- For \(x=3\): \(3\^{}3 - 3(3)\^{}2 + 3 = 27 - 27 + 3 = 3 \neq 0\).
- For \(x=-3\): \((-3)\^{}3 - 3(-3)\^{}2 + 3 = -27 - 27 + 3 = -51 \neq 0\).
None of the possible integer roots satisfy the equation. This confirms that there are no integer solutions.
Step 4: Final Answer:
There are 0 possible integer values of x that satisfy the equation.
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