Solutions

(a)

$$\begin{aligned} 2x^{2} - 3x + 4 &= 2(x^{2} - \frac{3x}{2} + 2) \\ &= 2(x^{2} -\frac{3x}{2} + \frac{32}{16} + (- \frac{23}{16} + \frac{23}{16})) \\ &= 2(x^{2} - \frac{3x}{2} + \frac{9}{16} + \frac{23}{16}) \\ &= 2((x - \frac{3}{4})^{2} + \frac{23}{16}) \\ &= 2(x - \frac{3}{4})^{2} + \frac{23}{8} \end{aligned}$$

This implies that the smallest value is $\frac{23}{8}$.

(b)

$$\begin{aligned} x^{2} - 3x + 2y^{2} + 4y + 2 &= x^{2} - 3x + (\frac{9}{4} - \frac{9}{4}) + 2y^{2} + 4y + 2 \\ &= {(x - \frac{3}{2})}^{2} - \frac{9}{4} + 2{(y + 1)}^{2} \end{aligned}$$

This implies that the smallest possible value is $\frac{-9}{4}$.

(c)

$$\begin{aligned} x^{2} + 4xy + 5y^{2} - 4x -6y + 7 &= x^{2} + 4xy +5y^{2} - (y^{2} - y^{2}) -4x - 6y + 7 \\ &= {(x + 2y)}^{2} + y^{2} - 4x - 6y + 7 \\ &= {(x + 2y)}^{2} -4x - 6y - (2y - 2y) + y^{2} + 7 \\ &= {(x + 2y)}^{2} -4(x + 2y) + 2y + y^{2} + 7 \\ &= {(x + 2y)}^{2} -4(x + 2y) + 7 + 2y + y^{2} \\ &= {(x + 2y)}^{2} -4(x + 2y) + (4 + 3) + 2y + y^{2} \\ &= {(x + 2y - 2)}^{2} + 3 + 2y + y^{2} \\ &= {(x + 2y - 2)}^{2} + y^{2} + 2y + 3 - (2 - 2) \\ &= {(x + 2y - 2)}^{2} + {(y + 1)}^{2} + 2 \end{aligned}$$

Thus, the smallest value is $2$.