Solution — 2019 Question 3 Quadratic Equations & Inequalities

(a)

Equating the two expressions for $y,$

\begin{matrix} 2 x^{2} + 6 x - 3 & = 11 x + 9 \\ 2 x^{2} - 5 x - 12 & = 0 \\ (2 x + 3) (x - 4) & = 0 \end{matrix}

When $x = - \frac{3}{2},$

\begin{matrix} y & = 11 (- \frac{3}{2}) + 9 \\ = - \frac{15}{2} \end{matrix}

When $x = 4,$

\begin{matrix} y & = 11 (4) + 9 \\ = 53 \end{matrix}

\begin{matrix} x & = - \frac{3}{2}, & y & = - \frac{15}{2} ∎ \\ or x & = 4, & y & = 53 ∎ \end{matrix}

(b)

From (a),

\begin{matrix} 2 x^{2} + 6 x - 3 & \leq 11 x + 9 \\ (2 x + 3) (x - 4) & \leq 0 \end{matrix}

$- \frac{3}{2} \leq x \leq 4 ∎$