1017SCG - Week 5

Foundation Mathematics

1017SCG

Week 5

Topics for Week 5

Solving Linear Equations
Solving Quadratic Equations
Solving Logarithmic Equations
Solving Exponential Equations

Solving Linear Equations

👉 $3x-7=9,\;\; 8y+2=9,\; 3n-1=-10$

Allowed: $\,x,\,$ constants ✅

Not allowed: $\,x^2,\,x^3,\, x^{1/2},\, \sin x,\, \log (x), \,3^x, \,$ ❌

Solving Linear Equations

Allowed: $\,x,\,$ constants ✅

Not allowed: $\,x^2,\,x^3,\, x^{1/2},\, \sin x,\, \log (x), \,3^x, \,$ ❌

This is a linear equation:

$3x+5 = 17$

$3x+5-5 = 17-5$

$3x = 12$

$\dfrac{3x}{3} = \dfrac{12}{3}$

$x = 4$

Solving Linear Equations

Example 1: Solve $\,7 n - 10 = 2n + 15$

$7n - 10 + 10$	$=$	$2n + 15 + 10$
$7n$	$=$	$2n + 25$
$7n - 2n$	$=$	$2n - 2n + 25$
$5n$	$=$	$25$
$\dfrac{5n}{5}$	$=$	$\dfrac{25}{5}$
$n$	$=$	$5$

Solving Linear Equations

Example 2: Solve $\,\dfrac{3 y +8}{2} = \dfrac{10y+1}{3}$

$\dfrac{3 y +8}{2}\times 2$	$=$	$ \dfrac{10y+1}{3} \times 2$
$3 y +8 $	$=$	$\dfrac{2(10y+1)}{3} $
$(3y+8)\times 3$	$=$	$\dfrac{2(10y+1)}{3} \times 3$
$3(3y+8)$	$=$	$2(10y+1)$
$9y+ 24$	$=$	$20 y + 2$

Solving Linear Equations

Example 2: Solve $\,\dfrac{3 y +8}{2} = \dfrac{10y+1}{3}$

$9y+ 24$	$=$	$20 y + 2$
$24-2$	$=$	$20y - 9y$
$22$	$=$	$11y$
$y$	$=$	$2$

Solving Quadratic Equations

$3x^2-7=9,\;\; y^2-y+1=0,\; n^2-4=-2$

Allowed: $\,x^2,\,x,\,$ constants ✅

Not allowed: $\,x^3,\,x^4,\, x^{1/2},\, \sin x,\, \log (x), \,3^x, \,$ etc. ❌

Solving Quadratic Equations

Example 1 (Factorising): Solve $\,x^2-6x = 0$

$x(x-6)$	$=$	$0$
$x=0$	or	$x -6=0$
		$x = 6$

Solving Quadratic Equations

Example 2 (Factorising): Solve $\,3x^2+6x = 0$

$3x(x+2)$	$=$	$ 0$
$3 x =0 $	or	$x+2 =0$
$x=0$	or	$x=-2$

Solving Quadratic Equations

Example 3 (Factorising): Solve $\,x^2 + 5 x +4 =0$

$(x+1)(x+4)$	$=$	$ 0$
$x+1 =0 $	or	$x+4 =0$
$x=-1$	or	$x=-4$

Solving Quadratic Equations

Example 4 (Factorising): Solve $\,2x^2 + 9 x +4 =0$

$(2x+1)(x+4)$	$=$	$ 0$
$2x+1 =0 $	or	$x+4 =0$
$2x=-1$	or	$x=-4$
$x=-\dfrac{1}{2}$

What if we cannot easily factorise?

Try this one for example:
$x^2+2x-2=0$

The Quadratic formula

If we have the quadratic equation \[ax^2+bx+c=0,\] then \[ x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a} \]

Note: Developed by Indian mathematician Brahmagupta (circa 6th century) and Śrīdhara (8th-9th century).

Derivation of the Quadratic formula 🤓

$ax^2+ bx+c $	$=$	$ 0$
$ax^2 + bx $	$=$	$ -c$
$4a$$ax^2 \,+$$\,4a$$bx $	$=$	$-$$\,4a$$c$
$4a^2x^2 + 4abx\, +$$\,b^2$	$=$	$ -4ac\,+$$\,b^2$
$\left(2ax+b\right)^2$	$=$	$ b^2-4ac$
$2ax+b$	$=$	$ \pm \sqrt{b^2-4ac}$
$2ax$	$=$	$ -b \pm \sqrt{b^2-4ac}$
$x$	$=$	$ \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

The Quadratic formula

If $ax^2+bx+c=0,$ then \[ x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a} \]

The symbol $\pm$ means that we get two solutions:

$ x=\dfrac{-b+ \sqrt{b^2-4ac}}{2a},\;\;\; $ $ x=\dfrac{-b- \sqrt{b^2-4ac}}{2a} $

Solving Quadratic Equations

Example 5: Solve $\,2x^2 -3 x -4 =0$

👉 $\;a = 2,\;\;$ $b = -3,\;\;$ $c = -4$

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\qquad \qquad\;\;$

$x = \dfrac{-(-3)\pm \sqrt{(-3)^2-4(2)(-4)}}{2(2)}$

$x = \dfrac{3\pm \sqrt{41}}{4}\qquad \qquad\qquad \quad$

Solving Quadratic Equations

Example 6: Solve $\,x^2 -6 x +9 =0$

👉 $\;a = 1,\;\;$ $b = -6,\;\;$ $c = 9$

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\qquad \qquad $

$x = \dfrac{-(-6)\pm \sqrt{(-6)^2-4(1)(9)}}{2(1)}$

$x = \dfrac{6\pm \sqrt{0}}{2}$ $= \dfrac{6}{2}$ $= 3 \qquad\quad \;\;\;$

👉 Note that we have only one solution‼

Solving Quadratic Equations

Example 6: Solve $\,x^2 -6 x +9 =0$

👉 Note that we have only one solution‼

We can solve this equation also by factorising!

$(x-3)(x-3)=0$

Then $\, x - 3 = 0 \,$ or $\,x-3=0$

This is just one equation $\, x - 3 = 0 $

Therefore $\, x =3$

Solving Quadratic Equations

🤔 What is the best strategy to solve $ax^2+ bx +c =0$?

$x^2-2x-2=0$	$x^2-7x+12=0$
$x=\dfrac{-(-2)\pm \sqrt{(-2)^2-4(1)(-2)}}{2(1)}$	$(x-3)(x-4)=0$
$x=1\pm \sqrt{3}$	$x=3, \; x = 4$

Solving Quadratic Equations

Final remark: When solving a quadratic equation, there are 3 possibilities:

There are 2 solutions: $x^2-2x-2=0$
There is 1 solution: $x^2-6x+9=0$
There are NO solutions: $x^2+1=0$

Geometric interpretation

Consider the cartesian coordinate system

Geometric interpretation

• 2 solutions: The parabola intersects the $x$-axis 2 times

Geometric interpretation

• 1 solution: The parabola intersects the $x$-axis once

Geometric interpretation

• 0 solutions: The parabola does not intersect the $x$-axis

Solving Logarithmic Equations

$\log_{10}(100)\, $ $= 2\;$ because $\,10^2 = 100$

$\log_{2}(8)\, $ $= 3\;$ because $\,2^3 = 8$

Solving Logarithmic Equations

Example 1: Solve $\,\log_{2}(x)=4 $

$2^4$	$=$	$ x$
$16$	$=$	$x$
$x$	$=$	$ 16$

Solving Logarithmic Equations

Example 2: Solve $\,\log_{5}(2x)=3 $

$5^3$	$=$	$ 2x$
$125$	$=$	$2x$
$\dfrac{125}{2}$	$=$	$ \dfrac{2x}{2}$
$x$	$=$	$ \dfrac{125}{2}$

Solving Logarithmic Equations

Example 3: Solve $\,5\log_{2}(3x-1)=10 $

$\dfrac{5\log_{2}(3x-1)}{5}$	$=$	$ \dfrac{10}{5}$
$\log_{2}(3x-1)$	$=$	$2$
$2^2$	$=$	$3x-1$
$4$	$=$	$ 3x-1$

Solving Logarithmic Equations

Example 3: Solve $\,5\log_{2}(3x-1)=10 $

$4$	$=$	$ 3x-1$
$4+1$	$=$	$ 3x-1+1$
$5$	$=$	$ 3x$
$\dfrac{5}{3}$	$=$	$ \dfrac{3x}{3}$
$x$	$=$	$ \dfrac{5}{3}$

Solving Exponential Equations

Example 1: Solve $\,5^x=100 $

$\log_{10}\left(5^x\right) $	$=$	$ \log_{10}(100)$
$x\log_{10}\left(5\right)$	$=$	$\log_{10}(100)$
$x\log_{10}\left(5\right)$	$=$	$2$
$\dfrac{x\log_{10}\left(5\right)}{\log_{10}(5)}$	$=$	$ \dfrac{2}{\log_{10}(5)}$
$x$	$=$	$ \dfrac{2}{\log_{10}(5)}$ $\approx 2.86$

Solving Logarithmic Equations

Example 1: Solve $\,5^x=100 $

👉 log₁₀(5ˣ)= log₁₀(100)

$\;\log_{5}\left(5^x\right) $	$=$	$ \log_{5}(100)$
$x\log_{5}\left(5\right)$	$=$	$\log_{5}(100)$
$x$	$=$	$\log_{5}(100)$
$x$	$=$	$\dfrac{\log_{10}\left(100\right)}{\log_{10}(5)}$ $=\dfrac{2}{\log_{10}(5)}$ $\approx 2.86$

Solving Logarithmic Equations

Example 2: Solve $\,e^{2x}=10 $

$\ln\left(e^{2x}\right)$	$=$	$ 10$
$2x\ln\left(e\right)$	$=$	$\ln\left(10\right)$
$2x$	$=$	$\ln\left(10\right)$
$\dfrac{2x}{2}$	$=$	$ \dfrac{\ln\left(10\right)}{2}$
$x$	$=$	$\dfrac{\ln\left(10\right)}{2}$ $\approx 1.15$

Solving Logarithmic Equations

Example 3: Solve $\,7e^{3x}=28 $

$\dfrac{7e^{3x}}{7}$	$=$	$ \dfrac{28}{7}$
$e^{3x}$	$=$	$4$
$\ln\left(e^{3x}\right)$	$=$	$\ln(4)$
$3x\ln\left(e\right)$	$=$	$\ln\left(4\right)$
$3x$	$=$	$\ln\left(4\right)$

Solving Logarithmic Equations

Example 3: Solve $\,7e^{3x}=28 $

$3x$	$=$	$\ln\left(4\right)$
$\dfrac{3x}{3}$	$=$	$ \dfrac{\ln\left(4\right)}{3}$
$x$	$=$	$ \dfrac{\ln\left(4\right)}{3}$ $\approx 0.46$

Why do we need to know how
to solve equations anyway?

$2x-7=4,\;\; 2x^2-x+1=0,\;\; \log_3(2x) = 6,\;\; 4^x+2 = 3 $

📊 Linear Regression

Predicting height from age: Suppose we collect data from children aged 2 to 13 and record their heights.

📊 Linear Regression

Predicting height from age:

The data points suggest a linear trend:

\[ h(x) = m x + 70.36 \]

📈 Linear Regression

Predicting height from age:

The data points suggest a linear trend:

\[ h(x) = m x + 70.36 \]

$ x $: age in years
$ h(x) $: predicted height in cm
Slope $ m $: average growth per year

This line is the line of best fit — found using linear regression.

🚀 Motion of Objects/Particles Affected by Gravity

$y(x) = \ds \frac{g}{2u^2} x^2+\frac{v}{u}x$

☕️ Law of Cooling & 🧫 Population Growth


$T(t)=\left(T_0-T_m\right) e^{-kt} + T_m$	$\ds P(t) = \frac{\theta P_0 e^{rt}}{\theta-P_0+P_0e^{rt}}$

In summary

Solving equations teaches logical thinking and is a powerful tool for predicting, designing, optimizing, and understanding the world around us.

An example of modelling using functions

The beauty of mathematics shows itself to patient followers.

- Maryam Mirzakhani

$R =13 + 3\left(\dfrac{1}{2} + \dfrac{1}{2} \sin \left(2\pi t + \dfrac{z}{3}\right)\right)^4$

$y = z -\abs{x}\sqrt{ \dfrac{20 - \abs{x}}{35} }$

$x^{2}+y^{2}+z^{2}=R$

Open in Desmos

\(7n - 10 + 10\)	\(=\)	\(2n + 15 + 10\)
\(7n\)	\(=\)	\(2n + 25\)
\(7n - 2n\)	\(=\)	\(2n - 2n + 25\)
\(5n\)	\(=\)	\(25\)
\(\dfrac{5n}{5}\)	\(=\)	\(\dfrac{25}{5}\)
\(n\)	\(=\)	\(5\)

\(\dfrac{3 y +8}{2}\times 2\)	\(=\)	\( \dfrac{10y+1}{3} \times 2\)
\(3 y +8 \)	\(=\)	\(\dfrac{2(10y+1)}{3} \)
\((3y+8)\times 3\)	\(=\)	\(\dfrac{2(10y+1)}{3} \times 3\)
\(3(3y+8)\)	\(=\)	\(2(10y+1)\)
\(9y+ 24\)	\(=\)	\(20 y + 2\)

\(9y+ 24\)	\(=\)	\(20 y + 2\)
\(24-2\)	\(=\)	\(20y - 9y\)
\(22\)	\(=\)	\(11y\)
\(y\)	\(=\)	\(2\)

\(x(x-6)\)	\(=\)	\(0\)
\(x=0\)	or	\(x -6=0\)
		\(x = 6\)

\(3x(x+2)\)	\(=\)	\( 0\)
\(3 x =0 \)	or	\(x+2 =0\)
\(x=0\)	or	\(x=-2\)

\((x+1)(x+4)\)	\(=\)	\( 0\)
\(x+1 =0 \)	or	\(x+4 =0\)
\(x=-1\)	or	\(x=-4\)

\((2x+1)(x+4)\)	\(=\)	\( 0\)
\(2x+1 =0 \)	or	\(x+4 =0\)
\(2x=-1\)	or	\(x=-4\)
\(x=-\dfrac{1}{2}\)

\(ax^2+ bx+c \)	\(=\)	\( 0\)
\(ax^2 + bx \)	\(=\)	\( -c\)
\(4a\)\(ax^2 \,+\)\(\,4a\)\(bx \)	\(=\)	\(-\)\(\,4a\)\(c\)
\(4a^2x^2 + 4abx\, +\)\(\,b^2\)	\(=\)	\( -4ac\,+\)\(\,b^2\)
\(\left(2ax+b\right)^2\)	\(=\)	\( b^2-4ac\)
\(2ax+b\)	\(=\)	\( \pm \sqrt{b^2-4ac}\)
\(2ax\)	\(=\)	\( -b \pm \sqrt{b^2-4ac}\)
\(x\)	\(=\)	\( \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\)

\(x^2-2x-2=0\)	\(x^2-7x+12=0\)
\(x=\dfrac{-(-2)\pm \sqrt{(-2)^2-4(1)(-2)}}{2(1)}\)	\((x-3)(x-4)=0\)
\(x=1\pm \sqrt{3}\)	\(x=3, \; x = 4\)

\(5^3\)	\(=\)	\( 2x\)
\(125\)	\(=\)	\(2x\)
\(\dfrac{125}{2}\)	\(=\)	\( \dfrac{2x}{2}\)
\(x\)	\(=\)	\( \dfrac{125}{2}\)

\(\dfrac{5\log_{2}(3x-1)}{5}\)	\(=\)	\( \dfrac{10}{5}\)
\(\log_{2}(3x-1)\)	\(=\)	\(2\)
\(2^2\)	\(=\)	\(3x-1\)
\(4\)	\(=\)	\( 3x-1\)

\(\log_{10}\left(5^x\right) \)	\(=\)	\( \log_{10}(100)\)
\(x\log_{10}\left(5\right)\)	\(=\)	\(\log_{10}(100)\)
\(x\log_{10}\left(5\right)\)	\(=\)	\(2\)
\(\dfrac{x\log_{10}\left(5\right)}{\log_{10}(5)}\)	\(=\)	\( \dfrac{2}{\log_{10}(5)}\)
\(x\)	\(=\)	\( \dfrac{2}{\log_{10}(5)}\) \(\approx 2.86\)

\(\;\log_{5}\left(5^x\right) \)	\(=\)	\( \log_{5}(100)\)
\(x\log_{5}\left(5\right)\)	\(=\)	\(\log_{5}(100)\)
\(x\)	\(=\)	\(\log_{5}(100)\)
\(x\)	\(=\)	\(\dfrac{\log_{10}\left(100\right)}{\log_{10}(5)}\) \(=\dfrac{2}{\log_{10}(5)}\) \(\approx 2.86\)

\(\ln\left(e^{2x}\right)\)	\(=\)	\( 10\)
\(2x\ln\left(e\right)\)	\(=\)	\(\ln\left(10\right)\)
\(2x\)	\(=\)	\(\ln\left(10\right)\)
\(\dfrac{2x}{2}\)	\(=\)	\( \dfrac{\ln\left(10\right)}{2}\)
\(x\)	\(=\)	\(\dfrac{\ln\left(10\right)}{2}\) \(\approx 1.15\)

\(\dfrac{7e^{3x}}{7}\)	\(=\)	\( \dfrac{28}{7}\)
\(e^{3x}\)	\(=\)	\(4\)
\(\ln\left(e^{3x}\right)\)	\(=\)	\(\ln(4)\)
\(3x\ln\left(e\right)\)	\(=\)	\(\ln\left(4\right)\)
\(3x\)	\(=\)	\(\ln\left(4\right)\)


\(T(t)=\left(T_0-T_m\right) e^{-kt} + T_m\)	$\ds P(t) = \frac{\theta P_0 e^{rt}}{\theta-P_0+P_0e^{rt}}$

Foundation Mathematics

Topics for Week 5

Solving Linear Equations

Solving Linear Equations

Solving Linear Equations

Solving Linear Equations

Solving Linear Equations

Solving Quadratic Equations

Solving Quadratic Equations

Solving Quadratic Equations

Solving Quadratic Equations

Solving Quadratic Equations

What if we cannot easily factorise?

The Quadratic formula

Derivation of the Quadratic formula 🤓

The Quadratic formula

Solving Quadratic Equations

Solving Quadratic Equations

Solving Quadratic Equations

Solving Quadratic Equations

Solving Quadratic Equations

Geometric interpretation

Geometric interpretation

Geometric interpretation

Geometric interpretation

Solving Logarithmic Equations

Solving Logarithmic Equations

Solving Logarithmic Equations

Solving Logarithmic Equations

Solving Logarithmic Equations

Solving Exponential Equations

Solving Logarithmic Equations

Solving Logarithmic Equations

Solving Logarithmic Equations

Solving Logarithmic Equations

Why do we need to know how to solve equations anyway?

📊 Linear Regression

📊 Linear Regression

📈 Linear Regression

🚀 Motion of Objects/Particles Affected by Gravity

☕️ Law of Cooling & 🧫 Population Growth

In summary

An example of modelling using functions

An example of modelling using functions

Why do we need to know how
to solve equations anyway?