Let integers a,b∈[−3,3] be such that a+b=0. Then the number of all possible ordered pairs (a, b), for which x+bx+1=1 and x+1ωω2ωz+ω21ω21z+ω=1, z∈C, where ω and ω2 are the roots of x2+x+1=0, is equal to:
Get the complete, step-by-step math solution for: "Let integers a, b [-3, 3] be such that a + b ≠ 0. Then the number of all possible ordered pairs (a, b), for which (x+1)/(x+b) = 1 and {ccc} x+1 & & ² ...". Powered by SolveForX AI math tutor.
Step-by-Step Solution
Step 1: Analyze the first equation
The first condition given is x+bx+a=1. This implies that the magnitude of the numerator is equal to the magnitude of the denominator, i.e., ∣x+a∣=∣x+b∣. Squaring both sides, we get ∣x+a∣2=∣x+b∣2. If x is a real number, this means (x+a)2=(x+b)2.
Step 2: Solve for x from the first equation
Expanding the squared terms, we get x2+2ax+a2=x2+2bx+b2. The x2 terms cancel out, leaving 2ax+a2=2bx+b2. Rearranging the terms to solve for x, we have 2ax−2bx=b2−a2, which simplifies to 2x(a−b)=b2−a2.
Step 3: Determine the value of x
We can factor the right side of the equation 2x(a−b)=b2−a2 as 2x(a−b)=(b−a)(b+a). Since b−a=−(a−b), we can write this as 2x(a−b)=−(a−b)(a+b). Given that a=b (otherwise a−b=0 and the equation would be 0=0, which doesn't determine x), we can divide both sides by (a-b).
Step 4: Solve for x
Dividing by (a-b) (assuming a=b), we get 2x=−(a+b). Therefore, x=−2a+b. Since a and b are integers, a+b is an integer. For x to be an integer, a+b must be an even number. This means a and b must have the same parity (both even or both odd).
Step 5: Analyze the second equation (determinant)
The second condition involves a determinant. We know that ω and ω2 are the roots of x2+x+1=0, which means 1+ω+ω2=0 and ω3=1. Let's simplify the determinant by applying row and column operations. Add C2 and C3 to C1.
Step 6: Simplify the determinant
Using the property 1+ω+ω2=0, the first column becomes x+1+ω+ω2=x+0=x, ω+z+ω2+1=z+(1+ω+ω2)=z+0=z, and ω2+1+z+ω=z+(1+ω+ω2)=z+0=z. So the determinant simplifies to xzzωz+ω21ω21z+ω=1. Now, subtract R2 from R3.
Step 7: Evaluate the determinant
Subtracting R2 from R3 gives a new third row: (z-z), (1−(z+ω2)), (z+ω−1). This simplifies to 0, (1−z−ω2), (z+ω−1). Now, expand the determinant along the first column. This gives x⋅((z+ω2)(z+ω−1)−(1)(1−z−ω2))−z⋅(ω(z+ω−1)−ω2(1−z−ω2))=1. This is a complex expression. Let's re-evaluate the determinant expansion. A simpler approach is to use C1→C1+C2+C3 first. We already did that. Let's expand the determinant directly from the simplified form: x⋅((z+ω2)(z+ω)−1)−ω⋅(z(z+ω)−z)+ω2⋅(z−z(z+ω2))=1. This is still complicated. Let's re-check the determinant properties. The determinant is of the form det(A). If we let C1→C1+C2+C3, we get x in the first column. Let's expand the determinant from the form xzzωz+ω21ω21z+ω=1. This is x((z+ω2)(z+ω)−1)−ω(z(z+ω)−z)+ω2(z−z(z+ω2))=1. This is x(z2+zω+zω2+ω3−1)−ω(z2+zω−z)+ω2(z−z2−zω2)=1. Since ω3=1 and 1+ω+ω2=0, we have zω+zω2=z(ω+ω2)=z(−1)=−z. So, x(z2−z+1−1)−ω(z2+zω−z)+ω2(z−z2−zω2)=1. This simplifies to x(z2−z)−ωz2−zω2+zω+zω2−z2ω2−zω4=1. This is getting too complex. Let's use the property that if x is real, then x must be 1. The determinant is of a special form. If x=1, then the determinant is 1ωω2ωz+ω21ω21z+ω. Add R2 and R3 to R1. This is not correct. Let's use the property that if x is real, then x=1. The determinant is x(z2+z(ω+ω2)+ω3−1)−ω(z(z+ω)−z)+ω2(z−z(z+ω2))=1. This simplifies to x(z2−z+1−1)−ω(z2+zω−z)+ω2(z−z2−zω2)=1. This is x(z2−z)−z2ω−zω2+zω+zω2−z2ω2−zω4=1. This is x(z2−z)−z2ω−zω2+zω+zω2−z2ω2−zω=1. This is x(z2−z)−z2ω−z2ω2=1. This is x(z2−z)−z2(ω+ω2)=1. This is x(z2−z)−z2(−1)=1. So x(z2−z)+z2=1. This is xz2−xz+z2=1. This equation must hold for some z∈C. If x=1, then z2−z+z2=1⟹2z2−z−1=0. This is a quadratic equation in z, which has solutions. So x=1 is a possible value. If x=1, then z2(x+1)−xz−1=0. For this quadratic in z to have solutions, the discriminant must be non-negative if z is real, but z can be complex. So x=1 is the only real value for x.
Step 8: Combine conditions for a and b
From the first equation, we found x=−2a+b. From the second equation, we found that x=1 is a possible value. Equating these, we get 1=−2a+b, which implies a+b=−2. We also know that a,b∈[−3,3] are integers, and a+b=0. The condition a+b=−2 satisfies a+b=0. Also, a+b=−2 is an even number, so a and b have the same parity, which is consistent with x being an integer.
Step 9: List possible pairs (a, b)
We need to find integer pairs (a, b) such that a,b∈[−3,3] and a+b=−2. Let's list the possible pairs:
If a=−3, then b=1. Pair: (−3,1).
If a=−2, then b=0. Pair: (−2,0).
If a=−1, then b=−1. Pair: (−1,−1).
If a=0, then b=−2. Pair: (0,−2).
If a=1, then b=−3. Pair: (1,−3).
If a=2, then b=−4, which is not in the range. So we stop here. All these pairs satisfy a+b=−2=0.
Step 10: Count the number of pairs
Counting the listed pairs, we have 5 possible ordered pairs (a, b) that satisfy all the given conditions.