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 zeroes \(\alpha, \beta\) is:
Sum of zeroes: \(\alpha + \beta = -b/a\)
Product of zeroes: \(\alpha\beta = c/a\)
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 \(a = p\), \(b = -14\), \(c = 8\).
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 \(\alpha = 2/p\) 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 \(8/p\):
\[ \frac{3}{p} = 1 \Rightarrow p = 3 \]
Step 4: Final Answer:
The value of \(p\) is 3.