3.3 Formulas and Surds
What is Rearranging?
Rearranging a formula means changing it so that a different variable becomes the subject (the variable on its own on one side of the equation).
Rearranging a formula means changing it so that a different variable becomes the subject (the variable on its own on one side of the equation).
⚡ The Golden Rule:
Whatever you do to one side of the equation, you must do to the other side.
Use inverse operations to isolate the variable you want.
Use inverse operations to isolate the variable you want.
💡 Inverse Operations:
• Addition ↔ Subtraction
• Multiplication ↔ Division
• Squaring ↔ Square rooting
• Powers ↔ Roots
• Multiplication ↔ Division
• Squaring ↔ Square rooting
• Powers ↔ Roots
Example 1: Make $x$ the subject of $y = 3x + 5$
Step 1: Subtract 5 from both sides
$$y - 5 = 3x$$
Step 2: Divide both sides by 3
$$\frac{y - 5}{3} = x$$
Answer: $x = \frac{y - 5}{3}$
$$y - 5 = 3x$$
Step 2: Divide both sides by 3
$$\frac{y - 5}{3} = x$$
Answer: $x = \frac{y - 5}{3}$
Example 2: Make $r$ the subject of $A = \pi r^2$
Step 1: Divide both sides by $\pi$
$$\frac{A}{\pi} = r^2$$
Step 2: Square root both sides
$$\sqrt{\frac{A}{\pi}} = r$$
Answer: $r = \sqrt{\frac{A}{\pi}}$
$$\frac{A}{\pi} = r^2$$
Step 2: Square root both sides
$$\sqrt{\frac{A}{\pi}} = r$$
Answer: $r = \sqrt{\frac{A}{\pi}}$
Example 3: Make $t$ the subject of $v = u + at$
Step 1: Subtract $u$ from both sides
$$v - u = at$$
Step 2: Divide both sides by $a$
$$\frac{v - u}{a} = t$$
Answer: $t = \frac{v - u}{a}$
$$v - u = at$$
Step 2: Divide both sides by $a$
$$\frac{v - u}{a} = t$$
Answer: $t = \frac{v - u}{a}$
🎯 Rearranging Practice
Real Life Uses:
• Physics: Rearranging $v = u + at$ to find time or acceleration
• Finance: Rearranging interest formulas to find principal or rate
• Engineering: Solving for unknown dimensions
• Physics: Rearranging $v = u + at$ to find time or acceleration
• Finance: Rearranging interest formulas to find principal or rate
• Engineering: Solving for unknown dimensions
Subject in the Denominator
When the variable you want is in the denominator of a fraction, multiply both sides to get it out.
When the variable is inside brackets, you may need to expand or deal with the bracket first.
When the variable you want is in the denominator of a fraction, multiply both sides to get it out.
Example 1: Make $x$ the subject of $y = \frac{5}{x}$
Step 1: Multiply both sides by $x$
$$xy = 5$$
Step 2: Divide both sides by $y$
$$x = \frac{5}{y}$$
Answer: $x = \frac{5}{y}$
$$xy = 5$$
Step 2: Divide both sides by $y$
$$x = \frac{5}{y}$$
Answer: $x = \frac{5}{y}$
Example 2: Make $R$ the subject of $I = \frac{V}{R}$
Subject Inside Brackets
Step 1: Multiply both sides by $R$
$$IR = V$$
Step 2: Divide both sides by $I$
$$R = \frac{V}{I}$$
Answer: $R = \frac{V}{I}$
$$IR = V$$
Step 2: Divide both sides by $I$
$$R = \frac{V}{I}$$
Answer: $R = \frac{V}{I}$
When the variable is inside brackets, you may need to expand or deal with the bracket first.
Example 3: Make $x$ the subject of $y = 3(x + 2)$
Method 1: Expand first
$y = 3x + 6$
$y - 6 = 3x$
$x = \frac{y - 6}{3}$
Method 2: Divide first
$\frac{y}{3} = x + 2$
$x = \frac{y}{3} - 2$
Both answers are equivalent!
$y = 3x + 6$
$y - 6 = 3x$
$x = \frac{y - 6}{3}$
Method 2: Divide first
$\frac{y}{3} = x + 2$
$x = \frac{y}{3} - 2$
Both answers are equivalent!
Example 4: Make $b$ the subject of $a = \frac{c + d}{b - 1}$
Step 1: Multiply both sides by $(b - 1)$
$$a(b - 1) = c + d$$
Step 2: Expand the brackets
$$ab - a = c + d$$
Step 3: Add $a$ to both sides
$$ab = c + d + a$$
Step 4: Divide by $a$
$$b = \frac{c + d + a}{a}$$
$$a(b - 1) = c + d$$
Step 2: Expand the brackets
$$ab - a = c + d$$
Step 3: Add $a$ to both sides
$$ab = c + d + a$$
Step 4: Divide by $a$
$$b = \frac{c + d + a}{a}$$
🎯 More Rearranging
Subject is Squared
When the variable is squared, take the square root at the end.
When the variable is under a square root, square both sides to remove it.
When the variable is squared, take the square root at the end.
Example 1: Make $v$ the subject of $E = \frac{1}{2}mv^2$
Step 1: Multiply both sides by 2
$$2E = mv^2$$
Step 2: Divide both sides by $m$
$$\frac{2E}{m} = v^2$$
Step 3: Square root both sides
$$v = \sqrt{\frac{2E}{m}}$$
$$2E = mv^2$$
Step 2: Divide both sides by $m$
$$\frac{2E}{m} = v^2$$
Step 3: Square root both sides
$$v = \sqrt{\frac{2E}{m}}$$
Example 2: Make $a$ the subject of $c^2 = a^2 + b^2$ (Pythagoras)
Subject is Under a Root
Step 1: Subtract $b^2$ from both sides
$$c^2 - b^2 = a^2$$
Step 2: Square root both sides
$$a = \sqrt{c^2 - b^2}$$
$$c^2 - b^2 = a^2$$
Step 2: Square root both sides
$$a = \sqrt{c^2 - b^2}$$
When the variable is under a square root, square both sides to remove it.
Example 3: Make $l$ the subject of $T = 2\pi\sqrt{\frac{l}{g}}$
Step 1: Divide both sides by $2\pi$
$$\frac{T}{2\pi} = \sqrt{\frac{l}{g}}$$
Step 2: Square both sides
$$\left(\frac{T}{2\pi}\right)^2 = \frac{l}{g}$$
Step 3: Multiply both sides by $g$
$$l = g\left(\frac{T}{2\pi}\right)^2 = \frac{gT^2}{4\pi^2}$$
$$\frac{T}{2\pi} = \sqrt{\frac{l}{g}}$$
Step 2: Square both sides
$$\left(\frac{T}{2\pi}\right)^2 = \frac{l}{g}$$
Step 3: Multiply both sides by $g$
$$l = g\left(\frac{T}{2\pi}\right)^2 = \frac{gT^2}{4\pi^2}$$
Example 4: Make $x$ the subject of $y = \sqrt{x + 3}$
Step 1: Square both sides
$$y^2 = x + 3$$
Step 2: Subtract 3 from both sides
$$x = y^2 - 3$$
$$y^2 = x + 3$$
Step 2: Subtract 3 from both sides
$$x = y^2 - 3$$
🎯 Even More Rearranging
The Factorising Method
When the subject appears more than once, collect all terms containing it on one side, then factorise to isolate it.
When the subject appears more than once, collect all terms containing it on one side, then factorise to isolate it.
⚡ The Method:
Step 1: Get all terms with the subject on one side
Step 2: Get all other terms on the other side
Step 3: Factorise out the subject
Step 4: Divide to isolate the subject
Step 2: Get all other terms on the other side
Step 3: Factorise out the subject
Step 4: Divide to isolate the subject
Example 1: Make $x$ the subject of $ax + b = cx + d$
Step 1: Get all $x$ terms on one side
$$ax - cx = d - b$$
Step 2: Factorise out $x$
$$x(a - c) = d - b$$
Step 3: Divide by $(a - c)$
$$x = \frac{d - b}{a - c}$$
$$ax - cx = d - b$$
Step 2: Factorise out $x$
$$x(a - c) = d - b$$
Step 3: Divide by $(a - c)$
$$x = \frac{d - b}{a - c}$$
Example 2: Make $y$ the subject of $\frac{y + a}{y + b} = c$
Step 1: Multiply both sides by $(y + b)$
$$y + a = c(y + b)$$
Step 2: Expand the brackets
$$y + a = cy + cb$$
Step 3: Collect $y$ terms on one side
$$y - cy = cb - a$$
Step 4: Factorise out $y$
$$y(1 - c) = cb - a$$
Step 5: Divide by $(1 - c)$
$$y = \frac{cb - a}{1 - c}$$
$$y + a = c(y + b)$$
Step 2: Expand the brackets
$$y + a = cy + cb$$
Step 3: Collect $y$ terms on one side
$$y - cy = cb - a$$
Step 4: Factorise out $y$
$$y(1 - c) = cb - a$$
Step 5: Divide by $(1 - c)$
$$y = \frac{cb - a}{1 - c}$$
Example 3: Make $x$ the subject of $y = \frac{x + 1}{x - 1}$
Step 1: Multiply both sides by $(x - 1)$
$$y(x - 1) = x + 1$$
Step 2: Expand
$$xy - y = x + 1$$
Step 3: Collect $x$ terms on one side
$$xy - x = 1 + y$$
Step 4: Factorise out $x$
$$x(y - 1) = 1 + y$$
Step 5: Divide by $(y - 1)$
$$x = \frac{1 + y}{y - 1}$$
$$y(x - 1) = x + 1$$
Step 2: Expand
$$xy - y = x + 1$$
Step 3: Collect $x$ terms on one side
$$xy - x = 1 + y$$
Step 4: Factorise out $x$
$$x(y - 1) = 1 + y$$
Step 5: Divide by $(y - 1)$
$$x = \frac{1 + y}{y - 1}$$
🎯 Rearranging with the subjec twice
What is a Surd?
A surd is an irrational number expressed as a root that cannot be simplified to a whole number or fraction. Examples include $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$.
To simplify a surd, find the largest perfect square factor.
A surd is an irrational number expressed as a root that cannot be simplified to a whole number or fraction. Examples include $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$.
💡 Surd or Not?
Surds: $\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{7}, \sqrt{11}...$
NOT Surds: $\sqrt{4} = 2$, $\sqrt{9} = 3$, $\sqrt{16} = 4$
If the number under the root is a perfect square, it's not a surd!
NOT Surds: $\sqrt{4} = 2$, $\sqrt{9} = 3$, $\sqrt{16} = 4$
If the number under the root is a perfect square, it's not a surd!
⚡ Rules for Surds:
Simplifying Surds
$$\sqrt{a} \times \sqrt{b} = \sqrt{ab}$$
$$\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$$
$$\sqrt{a^2} = a$$
$$(\sqrt{a})^2 = a$$
$$\sqrt{a} + \sqrt{a} = 2\sqrt{a}$$
To simplify a surd, find the largest perfect square factor.
Example 1: Simplify $\sqrt{50}$
Step 1: Find the largest perfect square factor of 50
$50 = 25 \times 2$ (25 is a perfect square)
Step 2: Split the root
$\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2}$
Step 3: Simplify
$= 5\sqrt{2}$
$50 = 25 \times 2$ (25 is a perfect square)
Step 2: Split the root
$\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2}$
Step 3: Simplify
$= 5\sqrt{2}$
Example 2: Simplify $\sqrt{72}$
Step 1: Find perfect square factors
$72 = 36 \times 2$ (36 is the largest perfect square factor)
Step 2: Simplify
$\sqrt{72} = \sqrt{36 \times 2} = \sqrt{36} \times \sqrt{2} = 6\sqrt{2}$
$72 = 36 \times 2$ (36 is the largest perfect square factor)
Step 2: Simplify
$\sqrt{72} = \sqrt{36 \times 2} = \sqrt{36} \times \sqrt{2} = 6\sqrt{2}$
Example 3: Simplify $\sqrt{48}$
$48 = 16 \times 3$
$\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$
$\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$
🎯 Surd Simplification Practice
Multiplying Surds
Example 1: Simplify $\sqrt{3} \times \sqrt{5}$
$$\sqrt{3} \times \sqrt{5} = \sqrt{3 \times 5} = \sqrt{15}$$
Example 2: Simplify $(3\sqrt{2})(4\sqrt{3})$
Step 1: Multiply the numbers: $3 \times 4 = 12$
Step 2: Multiply the surds: $\sqrt{2} \times \sqrt{3} = \sqrt{6}$
Answer: $(3\sqrt{2})(4\sqrt{3}) = 12\sqrt{6}$
Step 2: Multiply the surds: $\sqrt{2} \times \sqrt{3} = \sqrt{6}$
Answer: $(3\sqrt{2})(4\sqrt{3}) = 12\sqrt{6}$
Example 3: Simplify $(2\sqrt{3})^2$
$(2\sqrt{3})^2 = 2^2 \times (\sqrt{3})^2 = 4 \times 3 = 12$
Example 4: Expand $(1 + \sqrt{2})(3 - \sqrt{2})$
Dividing Surds
Using FOIL:
$= 1(3) + 1(-\sqrt{2}) + \sqrt{2}(3) + \sqrt{2}(-\sqrt{2})$
$= 3 - \sqrt{2} + 3\sqrt{2} - 2$
$= 1 + 2\sqrt{2}$
$= 1(3) + 1(-\sqrt{2}) + \sqrt{2}(3) + \sqrt{2}(-\sqrt{2})$
$= 3 - \sqrt{2} + 3\sqrt{2} - 2$
$= 1 + 2\sqrt{2}$
Example 5: Simplify $\frac{\sqrt{20}}{\sqrt{5}}$
$$\frac{\sqrt{20}}{\sqrt{5}} = \sqrt{\frac{20}{5}} = \sqrt{4} = 2$$
Example 6: Simplify $\frac{6\sqrt{10}}{2\sqrt{5}}$
Step 1: Divide the numbers: $\frac{6}{2} = 3$
Step 2: Divide the surds: $\frac{\sqrt{10}}{\sqrt{5}} = \sqrt{\frac{10}{5}} = \sqrt{2}$
Answer: $\frac{6\sqrt{10}}{2\sqrt{5}} = 3\sqrt{2}$
Step 2: Divide the surds: $\frac{\sqrt{10}}{\sqrt{5}} = \sqrt{\frac{10}{5}} = \sqrt{2}$
Answer: $\frac{6\sqrt{10}}{2\sqrt{5}} = 3\sqrt{2}$
🎯 Multiplying Surds
What is Rationalising?
Rationalising the denominator means removing the surd from the bottom of a fraction. This is considered the "proper" form for expressing answers.
When the denominator has two terms like $(a + \sqrt{b})$, multiply by the conjugate $(a - \sqrt{b})$.
Rationalising the denominator means removing the surd from the bottom of a fraction. This is considered the "proper" form for expressing answers.
⚡ Methods:
Simple Rationalising
Simple surd: Multiply top and bottom by the surd
$$\frac{a}{\sqrt{b}} \rightarrow \frac{a}{\sqrt{b}} \times \frac{\sqrt{b}}{\sqrt{b}} = \frac{a\sqrt{b}}{b}$$
With two terms (conjugate): Multiply by the conjugate
$$\frac{a}{b + \sqrt{c}} \rightarrow \times \frac{b - \sqrt{c}}{b - \sqrt{c}}$$
$$\frac{a}{\sqrt{b}} \rightarrow \frac{a}{\sqrt{b}} \times \frac{\sqrt{b}}{\sqrt{b}} = \frac{a\sqrt{b}}{b}$$
With two terms (conjugate): Multiply by the conjugate
$$\frac{a}{b + \sqrt{c}} \rightarrow \times \frac{b - \sqrt{c}}{b - \sqrt{c}}$$
Example 1: Rationalise $\frac{1}{\sqrt{5}}$
Step 1: Multiply top and bottom by $\sqrt{5}$
$$\frac{1}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{5}$$
$$\frac{1}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{5}$$
Example 2: Rationalise $\frac{6}{\sqrt{3}}$
$$\frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}$$
Example 3: Rationalise $\frac{4}{3\sqrt{2}}$
Rationalising with Conjugates
$$\frac{4}{3\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{4\sqrt{2}}{3 \times 2} = \frac{4\sqrt{2}}{6} = \frac{2\sqrt{2}}{3}$$
When the denominator has two terms like $(a + \sqrt{b})$, multiply by the conjugate $(a - \sqrt{b})$.
💡 Why does this work?
$(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$
This is the difference of two squares — the surd disappears!
This is the difference of two squares — the surd disappears!
Example 4: Rationalise $\frac{1}{2 + \sqrt{3}}$
Step 1: Multiply by the conjugate $\frac{2 - \sqrt{3}}{2 - \sqrt{3}}$
$$\frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}}$$
Step 2: Expand the denominator (DOTS)
$(2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1$
Answer: $\frac{2 - \sqrt{3}}{1} = 2 - \sqrt{3}$
$$\frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}}$$
Step 2: Expand the denominator (DOTS)
$(2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1$
Answer: $\frac{2 - \sqrt{3}}{1} = 2 - \sqrt{3}$
Example 5: Rationalise $\frac{3}{4 - \sqrt{2}}$
Step 1: Multiply by the conjugate $\frac{4 + \sqrt{2}}{4 + \sqrt{2}}$
$$\frac{3}{4 - \sqrt{2}} \times \frac{4 + \sqrt{2}}{4 + \sqrt{2}} = \frac{3(4 + \sqrt{2})}{16 - 2}$$
Step 2: Simplify
$$= \frac{12 + 3\sqrt{2}}{14}$$
$$\frac{3}{4 - \sqrt{2}} \times \frac{4 + \sqrt{2}}{4 + \sqrt{2}} = \frac{3(4 + \sqrt{2})}{16 - 2}$$
Step 2: Simplify
$$= \frac{12 + 3\sqrt{2}}{14}$$
🎯 Rationalise the Denominator
Real Life Uses:
• Engineering: Simplifying calculations with irrational measurements
• Physics: Working with constants like $\sqrt{2}$ in wave mechanics
• Architecture: Calculations involving diagonal measurements
• Computer Graphics: Normalising vectors
• Engineering: Simplifying calculations with irrational measurements
• Physics: Working with constants like $\sqrt{2}$ in wave mechanics
• Architecture: Calculations involving diagonal measurements
• Computer Graphics: Normalising vectors