Maths Demo (II)

• S4 Quadratic Function
  • Common Expressions of Roots
  • Completing the Square
  • Find Vertex
• S4 Complex Number
  • Power of i
  • Rationalize a Complex Number
• S4 Equations of Straight Line
  • Distance between a Point and a Line
• S4 Polynomials
  • Find Unknown Constants by Remainder Theorem
  • HCF and LCM of Polynomials
  • Long Division
  • Factorize Cubic Polynomial
  • Solve Cubic Equation
• S4 Exponentials and Logarithms
  • Solve Exponential Equation
  • Convert from Exponential to Logarithmic Form
  • Convert from Logarithmic to Exponential Form
• S4 Intersections of Straight Line and Parabola
  • Mid-point of Intersections
  • Horizontal Distance between Intersections
• S4 Variations
  • Joint Variation
  • Partial Variation

S4 Quadratic Function

Common Expressions of Roots

$α$ and $β$ are the roots of $2x^{2}+10x+6=0$. Find the value of

1. $α^{2}+β^{2}$

2. $(α-β)^{2}$

3. $α^{3}+β^{3}$

Solution

sum of root: $\begin{aligned} α+β &= -\dfrac{10}{2} \\ &= -5 \end{aligned}$

product of root: $\begin{aligned} αβ &= \dfrac{6}{2} \\ &= 3 \end{aligned}$

1. $\begin{aligned} α^{2}+β^{2} &= α^{2}+2αβ+β^{2}-2αβ \\ &= (α+β)^{2}-2αβ \\ &= (-5)^{2}-2(3) \\ &= 19 \end{aligned}$

2. $\begin{aligned} (α-β)^{2} &= α^{2}-2αβ+β^{2} \\ &= α^{2}+2αβ+β^{2}-4αβ \\ &= (α+β)^{2}-4αβ \\ &= (-5)^{2}-4(3) \\ &= 13 \end{aligned}$

3. $\begin{aligned} α^{3}+β^{3} &= (α+β)(α^{2}-αβ+β^{2}) \\ &= (-5)(α^{2}+β^{2}-3) \\ &= (-5)(19-3) && \leftarrow \text{by (a)} \\ &= -80 \end{aligned}$

Completing the Square

Complete the square for

1. $x^{2}-10x$

2. $2x^{2}-20x+5$

Solution

1. This is a simple version of completing the square.
   $\begin{aligned} x^{2}-10x &= x^{2}-2(5)x \\ &= \underline{x^{2}-2(5)x+5^{2}}-5^{2} \\ &= \underline{(x-5)^{2}}-25 \end{aligned}$

2. This is a general version of completing the square.
   $\begin{aligned} 2x^{2}-20x+5 &= 2\left[x^{2}-10x\right]+5 && \leftarrow \text{isolate out a simple version} \\ &= 2\left[(x-5)^{2}-25\right]+5 && \leftarrow \text{by (a)} \\ &= 2(x-5)^{2}-45 \end{aligned}$

Find Vertex

Find the coordinates of the vertex of $y=2x^{2}-20x+5$.

Method 1: By Formula

Let vertex = $(h, k)$

$\begin{aligned} h &= -\dfrac{b}{2a} \\ &= -\dfrac{-20}{2(2)} \\ &= 5 \end{aligned}$

Put $(5, k)$ into the equation:

$\begin{aligned} k &= 2(5)^{2}-20(5)+5 \\ &= -45 \end{aligned}$

∴     $(5, -45)$

Method 2: By Completing the Square (Slow!)

By completing the square, convert the equation to vertex form:

$\begin{aligned} 2x^{2}-20x+5 &= \ldots \\ &= 2(x-5)^{2}-45 \end{aligned}$

∴     $(5, -45)$ ← read off from vertex form $a(x-h)^{2}+k$

S4 Complex Number

Power of i

Simplify

1. $i^{4}$

2. $i^{23}$

3. $i^{-1}$

Solution

1. $\begin{aligned} i^{4} &= i^{2}i^{2} \\ &= (-1)(-1) \\ &= 1 \end{aligned}$

2. $\begin{aligned} i^{23} &= i^{4}i^{4}i^{4}i^{4}i^{4}i^{3} \\ &= i^{3} && \leftarrow i^{4}=1 \\ &= i^{2}i \\ &= -i \end{aligned}$

3. $\begin{aligned} i^{-1} &= \dfrac{1}{i}\times\dfrac{i}{i} \\ &= \dfrac{i}{i^{2}} \\ &= \dfrac{i}{-1} \\ &= -i \end{aligned}$

Rationalize a Complex Number

Express $\dfrac{1}{2+i}$ in the form of $a+bi$.

Solution

$\begin{aligned} \dfrac{1}{2+i}\times\dfrac{2-i}{2-i} &= \dfrac{2-i}{(2+i)(2-i)} \\ &= \dfrac{2-i}{2^{2}-i^{2}} \\ &= \dfrac{2-i}{4-(-1)} \\ &= \dfrac{2-i}{5} \\ &= \dfrac{2}{5}-\dfrac{1}{5}i \end{aligned}$

S4 Equations of Straight Line

Distance between a Point and a Line

Given a point $P(5, 3)$ and a straight line $L$: $y=2x+3$. Find the distance between $P$ and $L$.

Method 1: By Perpendicular Distance

The ‘distance’ between a point and a line means the shortest distance, which is also the perpendicular distance.

∴     distance between $P$ and $L$ = $PQ$, where $PQ\perp L$

Find the equation of $PQ$:

$\begin{aligned} m_{PQ}\times m_{L} &= -1 \\ m_{PQ}\times 2 &= -1 \\ m_{PQ} &= \dfrac{-1}{2} \end{aligned}$

∴     $\begin{aligned} \dfrac{y-3}{x-5} &= -0.5 \\ y &= -0.5x+5.5 \end{aligned}$

Find the coordinates of $Q$:

$\left\{ \begin{aligned} y &= 2x+3...L \\ y &= -0.5x+5.5...PQ \end{aligned} \right.$

∴     $Q=(1, 5)$ ← $Q$ is the intersection of $L$ and $PQ$

Find distance $PQ$:

$\begin{aligned} PQ &= \sqrt{(5-1)^{2}+(3-5)^{2}} \\ &= \sqrt{20} \end{aligned}$

Method 2: By Point-Line Distance Formula (Out of Syl.)

The distance $d$ between a point $(X, Y)$ and a straight line $ax+by+c=0$ is given by:

$d^{2}=\dfrac{\left[aX+bY+c\right]^{2}}{a^{2}+b^{2}}$

The distance between $P(5, 3)$ and $L$: $2x-y+3=0$ is given by:

$\begin{aligned} d^{2} &= \dfrac{\left[2(5)+(-1)(3)+3\right]^{2}}{2^{2}+(-1)^{2}} \\ &= 20 \\ d &= \sqrt{20} \end{aligned}$

S4 Polynomials

Find Unknown Constants by Remainder Theorem

Let $f(x)=2x^{3}+ax^{2}+bx-1$.

When $f(x)$ is divided by $x+2$, the remainder is −5. Also, $f(x)$ is divisible by $2x-1$. Find $a$ and $b$.

Solution

$\left\{ \begin{aligned} \text{f}(-2) &= -5 \\ \text{f}\left(\dfrac{1}{2}\right) &= 0 \end{aligned} \right. \to \left\{ \begin{aligned} 2(-2)^{3}+a(-2)^{2}+b(-2)-1 &= -5 \\ 2\left(\dfrac{1}{2}\right)^{3}+a\left(\dfrac{1}{2}\right)^{2}+b\left(\dfrac{1}{2}\right)-1 &= 0 \end{aligned} \right. \to \left\{ \begin{aligned} 4a-2b &= 12 \\ 0.25a+0.5b &= 0.75 \end{aligned} \right.$

Solving, $a=3$ and $b=0$. ← by calculator program

HCF and LCM of Polynomials

Find the HCF and LCM of $xy^{4}$, $x^{3}y^{2}$ and $x^{2}y^{3}z$.

Solution

$\begin{matrix} {} & {} & x & y^{4} & {} \\ {} & {} & x^{3} & y^{2} & {} \\ {} & {} & x^{2} & y^{3} & z \\ \text{HCF} & = & x & y^{2} & \\ \text{LCM} & = & x^{3} & y^{4} & z \\ \end{matrix}\quad \begin{matrix} {} \\ {} \\ {} \\ \text{take the lowest power} \\ \text{take the highest power} \\ \end{matrix}$

Long Division

Find the quotient and remainder for

1. $(2x^{3}+x^{2}-2x+10)\div (x+2)$

2. $(x^{4}+5x^{2}+x+3)\div (x^{2}+x)$

Solution

1. $\begin{array}{rrrr} 2x^{2} & -3x & +4 & \\ \hline \mathllap{x+2\;)\,}2x^{3} & +x^{2} & -2x & +10 \\ \underline{2x^{3}} & \underline{+4x^{2}} & & \\ & -3x^{2} & -2x & \\ & \underline{-3x^{2}} & \underline{-6x} & \\ & & 4x & +10 \\ & & \underline{4x} & \underline{+8} \\ & & & 2 \end{array}$
   ∴     quotient $=2x^{2}-3x+4$, remainder $=2$

2. $\begin{array}{rrrrr} x^{2} & -x & +6 & & \\ \hline \mathllap{x^{2}+x\;)\,}x^{4} & +0x^{3} & +5x^{2} & +x & +3 \\ \underline{x^{4}} & \underline{+x^{3}} & \underline{+0x^{2}} & & \\ & -x^{3} & +5x^{2} & +x & \\ & \underline{-x^{3}} & \underline{-x^{2}} & \underline{+0x} & \\ & & 6x^{2} & +x & +3 \\ & & \underline{6x^{2}} & \underline{+6x} & \underline{+0} \\ & & & -5x & +3 \end{array}$
   ←記得補零!
   ∴     quotient $=x^{2}-x+6$, remainder $=-5x+3$

Factorize Cubic Polynomial

Let $f(x)=x^{3}+4x^{2}+x-6$. Given that $x+2$ is a factor of $f(x)$, factorize $f(x)$.

Solution

∵     $\text{f}(x)=(x+2)Q(x)$ ← since $x+2$ is a factor

∴     $\begin{aligned} Q(x) &= \text{f}(x)\div (x+2) \\ &= x^{2}+2x-3 \end{aligned}$ ← by long division

∴     $\begin{aligned} \text{f}(x) &= (x+2)(x^{2}+2x-3) \\ &= (x+2)(x-1)(x+3) \end{aligned}$ ← by cross method

Solve Cubic Equation

Let $f(x)=x^{3}+4x^{2}+x-6$. Given that $x+2$ is a factor of $f(x)$, solve $f(x)=0$.

Solution

$\begin{aligned} \text{f}(x) &= 0 \\ (x+2)(x^{2}+2x-3) &= 0 \end{aligned}$ ← factorize $\text{f}(x)$

∴     $x+2=0$ or $x^{2}+2x-3=0$
$x=-2$, $x=1$ or $-3$

∴     $x=-2$, $1$ or $-3$

S4 Exponentials and Logarithms

Solve Exponential Equation

Solve

1. $2^{x}=3$

2. $2^{x-1}+2^{x+1}=20$

3. $4^{x}-3(2^{x})-10=0$

Solution

1. $\begin{aligned} 2^{x} &= 3 \\ x &= \log_{2}3 \\ &= 1.58 \end{aligned}$

2. $\begin{aligned} 2^{x-1}+2^{x+1} &= 20 \\ (2^{x})(2^{-1})+(2^{x})(2^{1}) &= 20 \\ (2^{x})(0.5)+(2^{x})(2) &= 20 \\ 2.5(2^{x}) &= 20 \\ 2^{x} &= 8 \\ x &= \log_{2}8 \\ &= 3 \end{aligned}$

3. $\begin{aligned} 4^{x}-3(2^{x})-10 &= 0 \\ (2^{x})^{2}-3(2^{x})-10 &= 0 && \leftarrow 4^{x}=(2^{2})^{x}=2^{2x}=(2^{x})^{2} \\ 2^{x} &= 5 \text{ or } -2 \text{ (rej.)} && \leftarrow 2^{x} \text{ is positive} \\ x &= \log_{2}5 \\ &= 2.32 \end{aligned}$

Convert from Exponential to Logarithmic Form

1. Express $y=2x^{3}$ in the form of $\log y=m\log x+c$.

2. Express $y=2(3^{x})$ in the form of $\log y=mx+c$.

Solution

Take log on both sides

1. $\begin{aligned} \log y &= \log(2x^{3}) \\ &= \log 2+\log x^{3} \\ &= \log 2+3\log x \\ &= 3\log x+\log 2 \end{aligned}$

2. $\begin{aligned} \log y &= \log(2(3^{x})) \\ &= \log 2+\log 3^{x} \\ &= \log 2+x\log 3 \\ &= (\log 3)x+\log 2 \end{aligned}$

Convert from Logarithmic to Exponential Form

1. Express $\log y=2\log x+3$ in the form of $y=ax^{b}$.

2. Express $\log y=2x+3$ in the form of $y=a(b^{x})$.

Method 1: Direct Method

(策略係將RHS combine做一個log)

1. $\begin{aligned} \log y &= 2\log x+3 \\ &= \log x^{2}+3 \\ &= \log x^{2}+\log 10^{3} \\ &= \log(x^{2}\times 10^{3}) \\ &= \log(1000x^{2}) \end{aligned}$
   ∴     $y=1000x^{2}$

2. $\begin{aligned} \log y &= 2x+3 \\ &= (\log 10^{2})x+\log 10^{3} \\ &= (\log 100)x+\log 1000 \\ &= \log 100^{x}+\log 1000 \\ &= \log(100^{x}\times 1000) \end{aligned}$
   ∴     $y=1000(100^{x})$

Method 2: Reverse Method

(策略係將個result寫返做logarithmic form然後Compare Term)

1. $y=ax^{b}$
   $\begin{aligned} \log y &= \log(ax^{b}) \\ &= b\log x+\log a \end{aligned}$
   Compare with $\log y=2\log x+3$,
   we have $b=2$ and $\begin{aligned} \log a &= 3 \\ a &= 10^{3} \\ &= 1000 \end{aligned}$
   ∴     $y=1000x^{2}$

2. $y=a(b^{x})$
   $\begin{aligned} \log y &= \log(a(b^{x})) \\ &= (\log b)x+\log a \end{aligned}$
   Comparing with $\log y=2x+3$,
   we have $\begin{aligned} \log b &= 2 \\ b &= 10^{2} \\ &= 100 \end{aligned}$ and $\begin{aligned} \log a &= 3 \\ a &= 10^{3} \\ &= 1000 \end{aligned}$
   ∴     $y=1000(100^{x})$

S4 Intersections of Straight Line and Parabola

Mid-point of Intersections

The straight line $y=2x+k$ cuts the parabola $y=x^{2}$ at $A$ and $B$.

Find the coordinates of the mid-point $M$ of $AB$ in terms of $k$.

Solution

Let $A=(α, ?)$, $B=(β, ?)$ and $M=(m, n)$

Put $y=2x+k$ into $y=x^{2}$:

$\begin{aligned} 2x+k &= x^{2} \\ 0 &= x^{2}-2x-k \end{aligned}$

∵     $α$ and $β$ are the roots of this equation

∴     $\begin{aligned} m &= \dfrac{α+β}{2} \\ &= \dfrac{-(-2)/1}{2} \\ &= 1 \end{aligned}$ ← sum of roots

Put $M=(1, n)$ into $y=2x+k$: ← $M$ lies on the straight line

$\begin{aligned} n &= 2(1)+k \\ &= 2+k \end{aligned}$

∴     $M=(1, 2+k)$

Horizontal Distance between Intersections

The straight line $y=2x+k$ cuts the parabola $y=x^{2}$ at $A$ and $B$.

Find the horizontal distance between $A$ and $B$ in terms of $k$.

Solution

Let $A=(α, ?)$ and $B=(β, ?)$.

The horizontal distance is $α-β$

Put $y=2x+k$ into $y=x^{2}$:

$\begin{aligned} 2x+k &= x^{2} \\ 0 &= x^{2}-2x-k \end{aligned}$

∵     $α$ and $β$ are the roots of this equation

∴     $\begin{aligned} α+β &= \dfrac{-(-2)}{1} \\ &= 2 \end{aligned}$ and $\begin{aligned} αβ &= \dfrac{-k}{1} \\ &= -k \end{aligned}$ ← sum and product of roots

∴     $\begin{aligned} (α-β)^{2} &= α^{2}-2αβ+β^{2} \\ &= α^{2}+2αβ+β^{2}-4αβ \\ &= (α+β)^{2}-4αβ \\ &= (2)^{2}-4(-k) \\ &= 4+4k \\ α-β &= \sqrt{4+4k} \end{aligned}$

S4 Variations

Joint Variation

$z$ varies directly as $x^{2}$ and inversely as $y$. When $x=2$ and $y=3$, we have $z=8$. Express $z$ in terms of $x$ and $y$.

Solution

Let $z=\dfrac{kx^{2}}{y}$ for some constant $k$

Put $x=2$, $y=3$ and $z=8$:

$\begin{aligned} 8 &= \dfrac{k(2)^{2}}{3} \\ k &= 6 \end{aligned}$

∴     $z=\dfrac{6x^{2}}{y}$

Partial Variation

$y$ partly varies as $x$ and partly varies as $x^{2}$. When $x=1$, $y=5$. When $x=2$, $y=16$. Express $y$ in terms of $x$.

Solution

Let $y=k_{1}x+k_{2}x^{2}$ for some constants $k_{1}$ and $k_{2}$

$\left\{ \begin{aligned} 5 &= k_{1}(1)+k_{2}(1)^{2} && ... \text{ when } x=1,\ y=5 \\ 16 &= k_{1}(2)+k_{2}(2)^{2} && ... \text{ when } x=2,\ y=16 \end{aligned} \right.$

∴     $k_{1}=2$, $k_{2}=3$ ← by calculator program

∴     $y=2x+3x^{2}$