Question

If $\frac{x^2 + y^2}{x+y} = 4$, then all possible values of (x - y) is given by

• A. [-2$\sqrt{2}$, 2$\sqrt{2}$]
• B. {-4, 4}
• C. [-4, 4]
• D. [-2, 2]

Answer 1

Answer: (C)

[-4, 4]

Answer 2

Let the given equation be $\frac{x^2 + y^2}{x+y} = 4$. This implies $x^2 + y^2 = 4(x+y)$, and we must have $x+y \ne 0$. Let $k = x-y$. Then $x = y+k$. Substituting $x$ in the equation: $(y+k)^2 + y^2 = 4((y+k)+y) \implies y^2 + 2yk + k^2 + y^2 = 4(2y+k) \implies 2y^2 + 2yk + k^2 = 8y + 4k$. Rearranging into a quadratic equation in $y$: $2y^2 + (2k-8)y + (k^2-4k) = 0$. For real solutions of $y$, the discriminant must be non-negative: $D = (2k-8)^2 - 4(2)(k^2-4k) \ge 0 \implies 4(k-4)^2 - 8(k^2-4k) \ge 0 \implies 4(k^2 - 8k + 16) - 8k^2 + 32k \ge 0 \implies 4k^2 - 32k + 64 - 8k^2 + 32k \ge 0 \implies -4k^2 + 64 \ge 0 \implies 64 \ge 4k^2 \implies 16 \ge k^2$. This implies $-4 \le k \le 4$. We need to check if $x+y=0$ can occur for any $k$ in this range. If $x+y=0$, then $y=-x$. Substituting $x-y=k$, we get $x-(-x)=k \implies 2x=k \implies x=k/2$. Then $y=-k/2$. If $x+y=0$, then $k/2 + (-k/2) = 0$. The original equation is undefined if $x+y=0$. If $y=-k/2$ is a root of $2y^2 + (2k-8)y + (k^2-4k) = 0$, then $2(-k/2)^2 + (2k-8)(-k/2) + (k^2-4k) = 0$, which simplifies to $k^2/2 = 0$, so $k=0$. When $k=0$, the quadratic for $y$ is $2y^2-8y=0 \implies 2y(y-4)=0$, giving $y=0$ or $y=4$. If $y=0$, $x=y+k=0$. Point $(0,0)$. $x+y=0$, so this solution is invalid. If $y=4$, $x=y+k=4$. Point $(4,4)$. $x+y=8 \ne 0$. This solution is valid, and $x-y=0$. Thus, $k=0$ is a possible value for $x-y$. The possible values of $x-y$ are in the interval $[-4, 4]$.