Question

If \( 8x^2 - 2kx + k = 0 \) is a quadratic equation in x, such that one of its roots is p times the other, and p, k are positive real numbers, then k equals

• A. \( (p + \frac{1}{p}) \)
• B. \( (\sqrt{p} + \frac{1}{\sqrt{p}})^2 \)
• C. \( 2(p + \frac{1}{p}) \)
• D. \( 2(\sqrt{p} + \frac{1}{\sqrt{p}})^2 \)

Hint:

When solving for a parameter in a quadratic equation based on a relationship between roots, the standard procedure is to use Vieta's formulas to create a system of equations and then eliminate the root variables to solve for the parameter.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem relates the coefficients of a quadratic equation to its roots using Vieta's formulas (sum and product of roots). We are given a relationship between the two roots, which allows us to solve for the coefficient k.
Step 2: Key Formula or Approach:
For a quadratic equation \( ax^2 + bx + c = 0 \) with roots \( \alpha \) and \( \beta \):

Sum of roots: \( \alpha + \beta = -b/a \)
Product of roots: \( \alpha \beta = c/a \)
Step 3: Detailed Explanation:
The given equation is \( 8x^2 - 2kx + k = 0 \). Let the roots be \( \alpha \) and \( \beta \). We are given that \( \beta = p \alpha \). Using Vieta's formulas:

Sum of roots: \[ \alpha + \beta = \alpha + p\alpha = \alpha(1+p) = -\frac{-2k}{8} = \frac{k}{4} \] \[ \alpha(1+p) = \frac{k}{4} \quad \text{(Equation 1)} \]
Product of roots: \[ \alpha \beta = \alpha(p\alpha) = p\alpha^2 = \frac{k}{8} \quad \text{(Equation 2)} \]
We have a system of two equations with two unknowns (\(\alpha\) and k). We want to find k. From Equation 1, we can express \( \alpha \) in terms of k: \[ \alpha = \frac{k}{4(1+p)} \] Now substitute this expression for \( \alpha \) into Equation 2: \[ p \left( \frac{k}{4(1+p)} \right)^2 = \frac{k}{8} \] \[ p \frac{k^2}{16(1+p)^2} = \frac{k}{8} \] Since k is a positive real number, \( k \neq 0 \), so we can divide both sides by k: \[ p \frac{k}{16(1+p)^2} = \frac{1}{8} \] Now, solve for k: \[ k = \frac{16(1+p)^2}{8p} = \frac{2(1+p)^2}{p} \] This is a correct expression for k, but it doesn't match the form of the options. We need to manipulate it algebraically. \[ k = \frac{2(1^2 + 2p + p^2)}{p} = 2 \left( \frac{1}{p} + \frac{2p}{p} + \frac{p^2}{p} \right) = 2 \left( \frac{1}{p} + 2 + p \right) \] Now let's expand the expression in option (4): \[ 2 \left( \sqrt{p} + \frac{1}{\sqrt{p}} \right)^2 = 2 \left( (\sqrt{p})^2 + 2(\sqrt{p})\left(\frac{1}{\sqrt{p}}\right) + \left(\frac{1}{\sqrt{p}}\right)^2 \right) \] \[ = 2 \left( p + 2 + \frac{1}{p} \right) \] This matches our derived expression for k.
Step 4: Final Answer:
The value of k is \( 2(\sqrt{p} + \frac{1}{\sqrt{p}})^2 \).