Answer :
[tex]\begin{gathered} xy=12 \\ x^2+y^2=25 \end{gathered}[/tex]
In order for us to determine the value of (x - y)(x - y), we have to solve for the value of x and y first. To do that, we can use the substitution method. Here are the steps.
1. Rewrite equation 1 xy = 12 in terms of "y" by dividing both sides of the equation by x.
[tex]\frac{xy}{x}=\frac{12}{x}\Rightarrow y=\frac{12}{x}[/tex]So, equation 1 becomes y = 12/x.
2. Replace the value of y in equation 2 with 12/x.
[tex]\begin{gathered} x^2+y^2=25 \\ x^2+(\frac{12}{x})^2=25 \end{gathered}[/tex]Then, solve for x.
[tex]x^2+\frac{144}{x^2}=25[/tex]Add the terms on the left side of the equation. Use x² as the GCD.
[tex]\frac{x^4+144}{x^2}=25[/tex]Cross multiply.
[tex]\begin{gathered} x^4+144=25x^2 \\ x^4-25x^2+144=0 \end{gathered}[/tex]Then, we factor the quartic polynomial. In order to factor, just answer the question "what are the factors of 144 that add to -25?".
The factors are -16 and -9. Hence, we can rewrite the equation as:
[tex]x^4-16x^2-9x^2+144=0[/tex]Then, divide the equation into two groups.
[tex](x^4-9x^2)-(16x^2-144)=0[/tex]Then, factor out x² in the first group and factor out 16 in the second group.
[tex]x^2(x^2-9)-16(x^2-9)=0[/tex]Since x² - 9 is a common factor on both groups, we can just rewrite it as:
[tex](x^2-16)(x^2-9)=0[/tex]Then, equate each factor to zero. Then, solve for x.
[tex]\begin{gathered} x^2-16=0 \\ x^2=16 \\ x=\pm4 \end{gathered}[/tex][tex]\begin{gathered} x^2-9=0 \\ x^2=9 \\ x=\pm3 \end{gathered}[/tex]Therefore, there are 4 possible values of x. These are +4, -4, +3, and -3.
Since xy = 12, then:
1. At x = 4, y = 3.
2. At x = -4, y = -3.
3. At x = 3, y = 4.
4. At x = -3, y = -4.
Let's plug the pairs of x and y in the expression (x - y)(x - y) and solve.
[tex]\begin{gathered} 1.\text{ }(4-3)(4-3)=(1)(1)=1 \\ 2.\text{ }(-4--3)(-4--3)=(-1)(-1)=1 \\ 3.\text{ }(3-4)(3-4)=(-1)(-1)=1 \\ 4.\text{ }(-3--4)(-3--4)=(1)(1)=1 \end{gathered}[/tex]On all pairs, the value of (x - y)(x- y) is 1. Hence, the answer is 1.