Question

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

• A. [-2√2, 2√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 $x^2+y^2 = 4(x+y)$. Let $S = x+y$ and $D = x-y$. We want to find the range of $D$. We know that $x^2+y^2 = \frac{(x+y)^2 + (x-y)^2}{2} = \frac{S^2+D^2}{2}$. Substituting this into the given equation: $\frac{S^2+D^2}{2} = 4S$ $S^2+D^2 = 8S$ $D^2 = 8S - S^2$ For $x$ and $y$ to be real numbers, the quadratic equation $t^2 - (x+y)t + xy = 0$ must have real roots. The roots of this equation are $x$ and $y$. We know $S = x+y$. Also, $(x+y)^2 = x^2+y^2+2xy$, so $S^2 = 4S + 2xy$. This gives $2xy = S^2 - 4S$. The discriminant of the quadratic $t^2 - St + xy = 0$ is $\Delta = S^2 - 4xy$. For real roots, $\Delta \ge 0$. $S^2 - 2(S^2 - 4S) \ge 0$ $S^2 - 2S^2 + 8S \ge 0$ $-S^2 + 8S \ge 0$ $S(8-S) \ge 0$ This inequality holds for $0 \le S \le 8$.

Now we need to find the range of $D^2 = 8S - S^2$ for $S \in [0, 8]$. Let $f(S) = 8S - S^2$. This is a quadratic function of $S$, representing a downward-opening parabola. The vertex of the parabola is at $S = -\frac{8}{2(-1)} = 4$. The maximum value of $f(S)$ occurs at $S=4$: $f(4) = 8(4) - (4)^2 = 32 - 16 = 16$. The minimum value of $f(S)$ occurs at the endpoints of the interval $[0, 8]$: $f(0) = 8(0) - 0^2 = 0$. $f(8) = 8(8) - 8^2 = 64 - 64 = 0$. So, the range of $D^2$ is $[0, 16]$. $0 \le D^2 \le 16$ Taking the square root, we get: $-4 \le D \le 4$ Thus, the possible values of $(x-y)$ are in the interval $[-4, 4]$.