Solutions

(a) Let's begin by testing the first number.

$$\begin{aligned} x^{2} + bx + c = 0 & \implies (\frac{-b + \sqrt{b^{2} -4c}}{2})^{2} + b\frac{-b + \sqrt{b^{2} -4c}}{2} + c = 0 \\ & \implies \frac{(-b + \sqrt{b^{2} -4c})^{2} }{4} + \frac{-b^{2} + b\sqrt{b^{2} -4c}}{2} + c = 0 \\ & \implies \frac{b^{2} - 2b\sqrt{b^{2} -4c} + (b^{2} -4c) }{4} + \frac{-2b^{2} + 2b\sqrt{b^{2} - 4c}}{4} + c = 0 \\ & \implies \frac{-4c}{4} + c = 0 \\ & \implies -c + c = 0 \\ & \implies 0 = 0 \end{aligned}$$

Moving on to the second number,

$$\begin{aligned} x^{2} + bx + c = 0 & \implies (\frac{-b - \sqrt{b^{2} -4c}}{2})^{2} + b\frac{-b - \sqrt{b^{2} -4c}}{2} + c = 0 \\ & \implies \frac{(-b - \sqrt{b^{2} -4c})^{2} }{4} + \frac{-b^{2} - b\sqrt{b^{2} -4c}}{2} + c = 0 \\ & \implies \frac{b^{2} + 2b\sqrt{b^{2} -4c} + (b^{2} -4c) }{4} + \frac{-2b^{2} - 2b\sqrt{b^{2} - 4c}}{4} + c = 0 \\ & \implies \frac{-4c}{4} + c = 0 \\ & \implies -c + c = 0 \\ & \implies 0 = 0 \end{aligned}$$

(b) We use a proof by contradiction.

$$\begin{aligned} x^{2} + bx + c = 0 & \implies x^{2} + bx + \frac{b^{2}}{4} - \frac{b^{2}}{4} + c = 0 \\ & \implies (x + \frac{b}{2})^{2} - \frac{b^{2}}{4} + c = 0 \\ & \implies (x + \frac{b}{2})^{2} = \frac{b^{2}}{4} - c \\ & \implies 4(x + \frac{b}{2})^{2} = b^{2} - 4c < 0 \\ & \implies 4(x + \frac{b}{2})^{2} < 0 \\ & \implies (x + \frac{b}{2})^{2} < 0 \end{aligned}$$

We thus end up with a contradiction since any number squared is greater or equal to 0.

(c)

$$\begin{aligned} b^{2} - 4c < 0 & \implies b^{2} < 4c \\ & \implies y^{2} < 4y^{2} && b = y \wedge c = y^{2} \\ & \implies x^{2} + yx + y^{2} > 0 && \text{Using b} \end{aligned}$$

(d) We first need to prove another proposition:

$$a^{2} < b \; \wedge \; b > 0 \implies -\sqrt{b} < a < \sqrt{b}$$$$\begin{aligned} a^{2} < b & \implies b > a^{2} \\ & \implies b - a^{2} > 0 \\ & \implies (\sqrt{b} - a)(\sqrt{b} + a) > 0 \\ & \implies [ \sqrt{b} - a > 0 \; \wedge \; \sqrt{b} + a > 0 ] \; \vee \; [ \sqrt{b} - a < 0 \; \wedge \; \sqrt{b} + a < 0 ] \\ & \implies [ \sqrt{b} > a \; \wedge \; \sqrt{a} > -b ] \; \vee \; [ \sqrt{b} < a \; \wedge \; \sqrt{b} < -a ] \\ & \implies -\sqrt{b} < a < \sqrt{b} \end{aligned}$$

Now, we can prove the main proposition.

$$\begin{aligned} b^{2} - 4c < 0 & \implies {(\alpha y)}^{2} - 4y^{2} < 0 && b = \alpha y \wedge c = y^{2} \\ & \implies \alpha^{2}y^{2} < 4y^{2} \\ & \implies \alpha^{2} < 4 \\ & \implies -2 < \alpha < 2 && \text{Proved prop.} \end{aligned}$$

(e) For $x^{2} + bx + c$, using the factorization done in (b), the quadratic is at least equal to $-\frac{b^{2}}{4} + c$.

For $ax^{2} + bx + c$, we factorize it like we factorized the
equation in (b), so as to create a perfect square and a constant
side term.

$$\begin{aligned} ax^{2} + bx + c &= a (x^{2} + \frac{bx}{a}) + c \\ &= a (x^{2} + \frac{bx}{a} + (\frac{b}{2a})^{2}) - (\frac{b}{2a})^{2}) ) + c \\ &= a ( (x + \frac{b}{2a})^{2} - \frac{b^{2}}{4a^{2}} ) + c \\ &= a(x + \frac{b}{2a})^{2} - a \frac{b^{2}}{4a^{2}} + c \\ &= a(x + \frac{b}{2a})^{2} - \frac{b^{2}}{4a} + c \end{aligned}$$

Thus, the smallest possible value of $ax^{2} + bx + c$ is $- \frac{b^{2}}{4a} + c$