Prove quadratic formula
Given $ax^2+bx+c=0$, then $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$Proof:$
\begin{array}{rl}
ax^2+bx+c &=0 \\
x^2+\frac{b}{a}x + \frac{c}{a} &=0 \\
x^2+\frac{b}{a}x &= - \frac{c}{a} \\
x^2+\frac{b}{a}x + \frac{b^2}{4a^2} &= \frac{b^2}{4a^2} - \frac{c}{a} \\
(x + \frac{b}{2a})^2 &= \frac{b^2-4ac}{4a^2} \\
x + \frac{b}{2a} &= \pm \sqrt{\frac{b^2-4ac}{4a^2}} \\
x &= -\frac{b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a} \\ \end{array}
$$ \therefore x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $$
$Proof:$
\begin{array}{rl}
ax^2+bx+c &=0 \\
x^2+\frac{b}{a}x + \frac{c}{a} &=0 \\
x^2+\frac{b}{a}x &= - \frac{c}{a} \\
x^2+\frac{b}{a}x + \frac{b^2}{4a^2} &= \frac{b^2}{4a^2} - \frac{c}{a} \\
(x + \frac{b}{2a})^2 &= \frac{b^2-4ac}{4a^2} \\
x + \frac{b}{2a} &= \pm \sqrt{\frac{b^2-4ac}{4a^2}} \\
x &= -\frac{b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a} \\ \end{array}
$$ \therefore x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $$
Proof: Angle in a semicircle is 90°
$AOB$ is a diameter of the circle, $C$ is a point on the circumference, forming the triangle $\triangle{ABC}$. The angle $\angle ACB$ is the 'angle in a semicircle'.
Radius $OC$ has been drawn, to form two isosceles triangles $\triangle{OAC}$ and $\triangle{OBC}$. They are isosceles as $OA$, $OB$ and $OC$ are all radiuses. So in $\triangle{OAC}$, let $\angle OAC=\alpha$ and in $\triangle{OBC}$, let $\angle OBC=\beta$ Hence $\angle OCA=\alpha$ (Angles in $\triangle{OAC}$) and $\angle OBC= \beta$ (Angles in $\triangle{OBC}$) By consider $\triangle{ABC}$, $\alpha +\alpha +\beta + \beta =180^{\circ}$ so $2\alpha+2\beta=180^{\circ}$ $\therefore \alpha+\beta =90^{\circ}$, $\angle ACB=90^{\circ}$ ie, 'Angle in a semicircle is a right angle' QED |