Question

Let $\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathit{a}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{b}\mathit{x}\mathbf{+}\mathit{c}\mathbf{\in }\mathbf{\square }\left[x\right]$with $\mathit{a}\mathbf{\ne }\mathbf{0}$. Prove that the roots of $\mathit{f}\left(x\right)$in $\mathbf{\square }$is $\frac{\mathbf{-}\mathbf{b}\mathbf{+}\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{a}\mathbf{c}}}{\mathbf{2}\mathbf{a}}$and $\frac{\mathbf{-}\mathbf{b}\mathbf{+}\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{ac}}}{\mathbf{2}\mathbf{a}}$. [Hint: Show that $\mathit{a}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{b}\mathit{x}\mathbf{+}\mathit{c}\mathbf{=}\mathbf{0}$is equivalent to ${\mathit{x}}^{\mathbf{2}}\mathbf{+}\left(\frac{b}{a}\right)\mathit{x}\mathbf{=}\frac{\mathbf{-}\mathbf{c}}{\mathbf{a}}$; then complete the square to find role="math" localid="1653653599624" $\mathit{x}$].

Short answer

The roots of $f\left(x\right)$in $\square$ is $\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $\frac{-b+\sqrt{b^2-4ac}}{2a}$ .

Step-by-step solution

Polynomial  fx

Consider the polynomial $\_$$f\left(x\right)=a{x}^2+bx+c\in \square \left[x\right]$with $a\ne 0$. Rewrite the polynomial as $f\left(x\right)=x^2+\left(\frac{b}{a}\right)x+\frac{c}{a}$.

Finding roots of  fx

Again, rewrite the equation as $x^2+\frac{b}{a}x=-\frac{c}{a}·$. Add ${\left(\frac{b}{2a}\right)}^2$ on both sides then,

$$\begin{gathered} x^2+\left(\frac{b}{a}\right)x+{\left(\frac{b}{2a}\right)}^2=-\frac{c}{a}+{\left(\frac{b}{2a}\right)}^2 \\ {\left(x+\frac{b}{2a}\right)}^2=\frac{b^2}{4a^2}-\frac{c}{a} \\ {\left(x+\frac{b}{2a}\right)}^2=\frac{b^2-4ac}{4a^2} \end{gathered}$$

Take square root on both sides, then simplify:

$$\begin{gathered} x+\frac{b}{2a}=\frac{\pm \sqrt{b^2-4ac}}{2a} \\ \frac{2ax+b}{2a}=\frac{\pm \sqrt{b^2-4ac}}{2a} \\ 2ax+b=\pm \sqrt{b^2-4ac} \\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \end{gathered}$$

Therefore, the required roots are $\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $\frac{-b-\sqrt{b^2-4ac}}{2a}$.