If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(x) = 2x^2 + 5x + k$ such that $\alpha^2 + \beta^2 + \alpha \beta = \dfrac{21}{4}$, find the value of $k$.
Step-by-Step Solution
Key Concept: Use $\alpha^2 + \beta^2 + \alpha \beta = (\alpha + \beta)^2 - \alpha \beta$ and substitute $\alpha+\beta = -5/2$, $\alpha\beta = k/2$.
Stepwise Solution:
From $p(x) = 2x^2 + 5x + k$, $\alpha + \beta = -\dfrac{5}{2}$ and $\alpha \beta = \dfrac{k}{2}$. [1.0 Mark]
We know $\alpha^2 + \beta^2 + \alpha \beta = (\alpha + \beta)^2 - 2\alpha\beta + \alpha\beta = (\alpha + \beta)^2 - \alpha\beta$. [0.5 Mark]
Substituting: $\left(-\dfrac{5}{2}\right)^2 - \dfrac{k}{2} = \dfrac{21}{4} \Rightarrow \dfrac{25}{4} - \dfrac{k}{2} = \dfrac{21}{4}$. [1.0 Mark]
$\dfrac{k}{2} = \dfrac{25}{4} - \dfrac{21}{4} = \dfrac{4}{4} = 1 \Rightarrow k = 2$. [0.5 Mark]
Marking Scheme:
• Finding sum and product of zeroes in terms of $k$: 1.0 Mark
• Writing identity $\alpha^2+\beta^2+\alpha\beta = (\alpha+\beta)^2 - \alpha\beta$: 0.5 Mark
• Substituting and setting up equation: 1.0 Mark
• Solving for $k = 2$: 0.5 Mark
Correct Answer: