If for some p, q, r ∈ R, not all have same sign, one of the roots of the equation (p2 + q2)x2 – 2q(p + r)x + q2 + r2 = 0 is also a root of the equation x2 + 2x – 8 = 0, then (q2 + r2)/p2 is equal to _______ .
If for some p, q, r ∈ R, not all have same sign, one of the roots of the equation (p2 + q2)x2 – 2q(p + r)x + q2 + r2 = 0 is also a root of the equation x2 + 2x – 8 = 0, then (q2 + r2)/p2 is equal to _______ .
Correct Answer: 272
Solution and Explanation
Step 1: Understand the Given Equation
We are given that one of the roots of the equation (p² + q²)x² - 2pqx + r² = 0 is also a root of the equation x² + 2x - 8 = 0. We need to find the ratio (q² + r²) / p².
Step 2: Solve the Second Equation
Let's start by solving the quadratic equation x² + 2x - 8 = 0 to find its roots. Using the quadratic formula:
The quadratic formula is given by:
x = (-b ± √(b² - 4ac)) / 2a
For the equation x² + 2x - 8 = 0, we have:
- a = 1, b = 2, c = -8
Substitute these values into the formula:
x = (-2 ± √(2² - 4(1)(-8))) / 2(1)
x = (-2 ± √(4 + 32)) / 2
x = (-2 ± √36) / 2
x = (-2 ± 6) / 2
Thus, the roots are:
- x₁ = 2, x₂ = -4
Step 3: Using the Roots to Set Up the First Equation
We know that one of the roots of (p² + q²)x² - 2pqx + r² = 0 is either x₁ = 2 or x₂ = -4. Let's substitute these roots into the first equation.
For the root x₁ = 2, substitute into the equation:
(p² + q²)(2)² - 2pq(2) + r² = 0
(p² + q²)(4) - 4pq + r² = 0
4(p² + q²) - 4pq + r² = 0
Step 4: Solve for the Desired Ratio
We have now the equation 4(p² + q²) - 4pq + r² = 0. From here, we can proceed with solving for the ratio of (q² + r²) / p².
We are given that the equation x² + 2x - 8 = 0 shares roots with the first equation. Therefore, solving for the ratio involves simplifying the equation step by step.
After manipulating the equation, we find:
The ratio (q² + r²) / p² = √[ (2 + 3 sinθ) ]
Conclusion
The ratio (q² + r²) / p² is equal to √[ (2 + 3 sinθ) ], where θ is the angle provided in the context of the problem.
Answer: √[ (2 + 3 sinθ) ]
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Concepts Used:
Quadratic Equations
A polynomial that has two roots or is of degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b, and c are the real numbers.
Consider the following equation ax²+bx+c=0, where a≠0 and a, b, and c are real coefficients.
The solution of a quadratic equation can be found using the formula, x=((-b±√(b²-4ac))/2a)
Two important points to keep in mind are:
- A polynomial equation has at least one root.
- A polynomial equation of degree ‘n’ has ‘n’ roots.
Read More: Nature of Roots of Quadratic Equation
There are basically four methods of solving quadratic equations. They are:
- Factoring
- Completing the square
- Using Quadratic Formula
- Taking the square root