The zeroes of a polynomial $x^2 + px + q$ are twice the zeroes of the polynomial $4x^2 - 5x - 6$. The value of $p$ is :
Generated by claude-sonnet-4-6 · 2026-06-15 10:36 · grounding rag
Model Answer
(c) −5
For $4x^2 - 5x - 6$: sum of zeroes $= \dfrac{5}{4}$. Zeroes of $x^2 + px + q$ are twice these, so their sum $= \dfrac{5}{2}$. Since sum $= \dfrac{-p}{1}$, we get $p = -5$.
Source: Chapter 2, Section 2.3
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Explanation
- First find the sum of zeroes of $4x^2-5x-6$ using $\alpha+\beta = \frac{-b}{a} = \frac{5}{4}$.
- The new zeroes are $2\alpha$ and $2\beta$, so their sum $= 2 \times \frac{5}{4} = \frac{5}{2}$.
- For $x^2+px+q$, sum of zeroes $= \frac{-p}{1} = \frac{5}{2}$, giving $p = -5$.
- A common mistake is forgetting to double the sum; another is sign error when equating $-p = \frac{5}{2}$.