Key Concepts and Formulas

• Roots of a Quadratic Equation: For a quadratic equation $ax^2 + bx + c = 0$, the roots are given by $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$.
• Sum of Squares Principle: For any real numbers $A$ and $B$, if $A^2 + B^2 = 0$, then it must be that $A = 0$ and $B = 0$.

Step-by-Step Solution

Step 1: Analyze the given quadratic equations and find the roots of the simpler equation.

We are given two quadratic equations:

1. $(p^{2}+q^{2}) x^{2}-2 q(p+r) x+q^{2}+r^{2}=0 \quad \ldots(1)$
2. $x^{2}+2 x-8=0 \quad \ldots(2)$

First, let's find the roots of equation (2) by factoring: $x^2 + 2x - 8 = 0$ $(x+4)(x-2) = 0$ Therefore, the roots of equation (2) are $x = 2$ and $x = -4$.

Step 2: Analyze equation (1) and determine the possible common root.

The coefficients of equation (1) are: $A = p^2+q^2$ $B = -2q(p+r)$ $C = q^2+r^2$

Since $p, q, r$ are real numbers, $p^2 \ge 0$, $q^2 \ge 0$, $r^2 \ge 0$. Also, $p^2 + q^2 > 0$ because otherwise equation (1) wouldn't be quadratic. Similarly, $q^2+r^2 \ge 0$.

The problem states that one of the roots of equation (1) is also a root of equation (2). Therefore, the common root can be either $2$ or $-4$.

Step 3: Assume the common root is 2 and derive the relationships between p, q, and r.

If $x=2$ is a root of equation (1), then substituting $x=2$ into equation (1) yields: $(p^2+q^2)(2)^2 - 2q(p+r)(2) + (q^2+r^2) = 0$ $4(p^2+q^2) - 4q(p+r) + q^2+r^2 = 0$ $4p^2 + 4q^2 - 4pq - 4qr + q^2 + r^2 = 0$ $4p^2 - 4pq + q^2 + 4q^2 - 4qr + r^2 = 0$ $(2p-q)^2 + (2q-r)^2 = 0$

Applying the sum of squares principle, this implies: $2p-q = 0 \quad \Rightarrow \quad q = 2p \quad \ldots(3)$ $2q-r = 0 \quad \Rightarrow \quad r = 2q \quad \ldots(4)$ Substituting (3) into (4): $r = 2(2p) = 4p$ So, if $x=2$ is the common root, then $q=2p$ and $r=4p$.

Step 4: Check if the condition "not all have the same sign" is satisfied when the common root is 2.

If $p > 0$, then $q = 2p > 0$ and $r = 4p > 0$. All three are positive. If $p < 0$, then $q = 2p < 0$ and $r = 4p < 0$. All three are negative. If $p=0$, then $q=0$ and $r=0$. In all cases, $p, q, r$ have the same sign or are all zero. This contradicts the condition that $p, q, r$ are not all of the same sign. Therefore, $x=2$ cannot be the common root.

Step 5: Assume the common root is -4 and derive the relationships between p, q, and r.

If $x=-4$ is a root of equation (1), then substituting $x=-4$ into equation (1) yields: $(p^2+q^2)(-4)^2 - 2q(p+r)(-4) + (q^2+r^2) = 0$ $16(p^2+q^2) + 8q(p+r) + q^2+r^2 = 0$ $16p^2 + 16q^2 + 8pq + 8qr + q^2 + r^2 = 0$ $16p^2 + 8pq + q^2 + 16q^2 + 8qr + r^2 = 0$ $(4p+q)^2 + (4q+r)^2 = 0$

Applying the sum of squares principle, this implies: $4p+q = 0 \quad \Rightarrow \quad q = -4p \quad \ldots(5)$ $4q+r = 0 \quad \Rightarrow \quad r = -4q \quad \ldots(6)$ Substituting (5) into (6): $r = -4(-4p) = 16p$ So, if $x=-4$ is the common root, then $q=-4p$ and $r=16p$.

Step 6: Check if the condition "not all have the same sign" is satisfied when the common root is -4.

If $p > 0$, then $q = -4p < 0$ and $r = 16p > 0$. Not all three have the same sign. If $p < 0$, then $q = -4p > 0$ and $r = 16p < 0$. Not all three have the same sign. Therefore, the common root must be $x=-4$.

Step 7: Calculate the desired expression.

We have found the relationships $q = -4p$ and $r = 16p$. Now we need to calculate the value of $\frac{q^2+r^2}{p^2}$. Substitute the expressions for $q$ and $r$ in terms of $p$: $\frac{q^2+r^2}{p^2} = \frac{(-4p)^2 + (16p)^2}{p^2}$ $= \frac{16p^2 + 256p^2}{p^2}$ $= \frac{272p^2}{p^2}$ Since $p \neq 0$, we can cancel $p^2$: $= 272$

Common Mistakes & Tips

• Carefully consider all conditions: The condition "not all have the same sign" is crucial for identifying the correct common root.
• Sum of Squares: Recognize and utilize the sum of squares principle effectively.
• Sign Errors: Double-check signs when manipulating equations.

Summary

By analyzing the given quadratic equations and using the condition that the coefficients $p, q, r$ are not all of the same sign, we determined that the common root is -4. This allowed us to find the relationships $q = -4p$ and $r = 16p$. Finally, we calculated the value of the expression $\frac{q^2+r^2}{p^2}$, which equals 272.

Final Answer The final answer is \boxed{272}.