Category Archives: Solving Equations

March 31, 2025 · 7:14 am

Deriving the Quadratic Equation formula

My year 10 students have been learning how to complete the square with the idea of then deriving the quadratic equation formula.

The general equation for a quadratic is $y=ax^2+bx+c$

Completing the square,

$\begin{equation*}ax^2+bx+c\end{equation}$

Factorise out the leading coefficient (i.e. $a$ )

$\begin{equation*}a(x^2+\frac{bx}{a}+\frac{c}{a})\end{equation}$

Half the second term (i.e $\frac{b}{a}$ ) and subtract the square of the second term.

$\begin{equation*}a((x+\frac{b}{2a})^2-(\frac{b}{2a})^2+\frac{c}{a})\end{equation}$

$\begin{equation*}a((x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{c}{a})\end{equation}$

Simplify

$\begin{equation*}a((x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{4ac}{4a^2})\end{equation}$

$\begin{equation*}a((x+\frac{b}{2a})^2+\frac{-b^2+4ac}{4a^2})\end{equation}$

$\begin{equation*}a(x+\frac{b}{2a})^2+\frac{-b^2+4ac}{4a}\end{equation}$

Now let’s solve

$\begin{equation*}a(x+\frac{b}{2a})^2+\frac{-b^2+4ac}{4a}=0\end{equation}$

$\begin{equation*}a(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a}\end{equation}$

$\begin{equation*}(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}\end{equation}$

$\begin{equation*}(x+\frac{b}{2a})=\pm \sqrt{\frac{b^2-4ac}{4a^2}}\end{equation}$

$\begin{equation*}(x+\frac{b}{2a})=\frac{\pm \sqrt{b^2-4ac}}{2a}\end{equation}$

$\begin{equation*}x=-\frac{b}{2a}\frac{\pm \sqrt{b^2-4ac}}{2a}\end{equation}$

Which is the quadratic equation formula.

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Filed under Algebra, Quadratic, Quadratics, Solving, Solving, Solving Equations

Tagged as completing the square, quadratic equation formula

December 3, 2024 · 6:54 am

Puzzle Page 1

If $\frac{a+b+2c}{a+b-c}=\frac{31}{15}$ , what does $\frac{a+b}{c}$ equal?

$\begin{equation*}15(a+n+2c)=31(a+b-c)\end{equation}$

$\begin{equation*}15a+15b+30c=31a+31b-31c)\end{equation}$

$\begin{equation*}61c=16a+16b)\end{equation}$

$\begin{equation*}\frac{61}{16}=\frac{a+b}{c}\end{equation}$

Two positive numbers are such that their difference, their sum, and their product are in the ratio $2:5:21$ . What is the smaller of the two numbers?

Let $x$ and $y$ be the two numbers. Then

(1) $\begin{equation*}x-y=2k\end{equation*}$

(2) $\begin{equation*}x+y=5k\end{equation*}$

(3) $\begin{equation*}xy=21k\end{equation*}$

Add equation $1$ and $2$ together to eliminate the $y$

$\begin{equation*}2x=7k\end{equation}$

(4) $\begin{equation*}x=\frac{7k}{2}\end{equation*}$

From $2$ $=5k-x$ , substitute for $y$ into equation $3$ .

(5) $\begin{equation*}x(5k-x)=21k\end{equation*}$

Substitute $x=\frac{7k}{2}$ into equation $5$ .

$\begin{equation*}\frac{7k}{2}(5k-\frac{7k}{2})=21k\end{equation}$

$\begin{equation*}\frac{35k^2}{2}-\frac{49k^2}{4}=21k\end{equation}$

$\begin{equation*}\frac{70k^2}{4}-\frac{49k^2}{4}=\frac{84k}{4}\end{equation}$

$\begin{equation*}21k^2-84k=0\end{equation}$

$\begin{equation*}21k(k-4)=0\end{equation}$

Hence, $k=0$ or $k=4$ .

When $k=4$ , $x=\frac{7\times 4}{2}=14$ and $y=5\times 4-14=6$

Therefore the smaller number is $6$ .

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Filed under Algebra, Puzzles, Ratio, Solving Equations

Tagged as problem solving, ratio, Solving Equations