Question

Find the value of p, for which one zero of the quadratic polynomial \(px^2 - 14x + 8\) is 6 times the other.

Hint:

Always solve the "sum" equation for the variable zero (\(\alpha\)) first, then plug that result into the "product" equation to find the unknown coefficient.

Answer 1

Step 1: Understanding the Concept:
For a polynomial \( ax^2 + bx + c \), the relationship between the zeroes \( \alpha \) and \( \beta \) is:
Sum of zeroes: \( \alpha + \beta = -\frac{b}{a} \)
Product of zeroes: \( \alpha\beta = \frac{c}{a} \)
Step 2: Key Formula or Approach:
Let the zeroes be \( \alpha \) and \( 6\alpha \).
Step 3: Detailed Explanation:
Given polynomial: \( px^2 - 14x + 8 \).
Here \( a = p \), \( b = -14 \), and \( c = 8 \).
1. Sum of zeroes:
\[ \alpha + 6\alpha = -\frac{-14}{p} \] \[ 7\alpha = \frac{14}{p} \Rightarrow \alpha = \frac{2}{p} \] 2. Product of zeroes:
\[ \alpha \cdot (6\alpha) = \frac{8}{p} \] \[ 6\alpha^2 = \frac{8}{p} \] 3. Substitute \( \alpha = \frac{2}{p} \) into the product equation:
\[ 6 \left( \frac{2}{p} \right)^2 = \frac{8}{p} \] \[ 6 \left( \frac{4}{p^2} \right) = \frac{8}{p} \] \[ \frac{24}{p^2} = \frac{8}{p} \] Since \( p \neq 0 \), we can divide by \( \frac{8}{p} \):
\[ \frac{3}{p} = 1 \Rightarrow p = 3 \] Step 4: Final Answer:
The value of \( p \) is 3.