IB Mathematics Analysis and Approaches Topic SL…

Why This Exists: Not Every Equation Rearranges To x = Something

Up to now, almost every equation you have solved could be rearranged step by step until $x$ sat alone on one side.
A linear equation, a factorised quadratic, or a simple exponential all give up their answer to pure algebra.
But most equations that describe the real world are not that polite: $e^x = \sin x$ cannot be rearranged to $x = \dots$ by any amount of algebra, because $x$ is trapped inside two different kinds of function at once.
So mathematicians keep two toolkits: an analytic one for equations that a method can crack, and a graphical or technology one for equations that no method can.
This subtopic is about owning both toolkits and, just as importantly, knowing on sight which one a given equation needs.

Analogy

Think of an equation like $g(x) = h(x)$ as a question: at what input do these two machines produce the same output?
Drawn as two curves, the answer is simply where they cross, so solving an equation and finding an intersections of graphs are the same act seen two ways.

Every Solution Is Either A Crossing Or A Zero

There are two pictures behind everything in this subtopic, and both turn an equation into something you can see.
Picture 1: A crossing (two curves)
1. To solve $g(x) = h(x)$, draw $y = g(x)$ and $y = h(x)$ on the same axes.
2. Each point where the curves meet has one $x$-value, and that $x$-value is a solution of the equation.
Picture 2: A zero (one curve)
1. Move everything to one side so the equation reads $f(x) = 0$, then draw the single curve $y = f(x)$.
2. Every place the curve cuts the $x$-axis is a solution, because there $y = f(x) = 0$.
The two pictures are interchangeable: $g(x) = h(x)$ becomes $g(x) - h(x) = 0$, which swaps a crossing for a zero of $f(x) = g(x) - h(x)$.

Tip

Whether you hunt for a crossing or a zero, you are looking for the same $x$-values; choose whichever picture is easier to draw.

Note

An analytic method finds these $x$-values exactly, as whole numbers, fractions, surds, or logarithms.
A graphical or technology method finds them as decimals read off a picture, usually to three significant figures.
Both are finding the same solutions; they differ only in whether the answer comes out exact or approximate.

Analytic Methods: Use Them Whenever A Method Exists

An analytic solution reaches the answer by exact algebra alone, and it is always preferred when a method is available, because the answer is exact.
Three moves cover almost every analytically solvable equation in the course.
Move 1: Factorise, then use the zero product property
1. If you can write $f(x) = 0$ as a product of factors, then a product is zero only when a factor is zero, which you meet fully when factorising and the quadratic formula.
Move 2: Rearrange and isolate, then undo
1. Get the awkward part alone on one side, then apply its inverse, for example taking $\ln$ of both sides to undo an exponential, as with exponential and logarithmic equations.
Move 3: Substitute to reveal a hidden quadratic
1. Some equations are secretly quadratics wearing a disguise, and a single substitution unmasks them.

Example

Consider $e^{2x} - 5e^x + 4 = 0$, which looks nothing like a quadratic at first glance.
Notice that $e^{2x} = \left(e^x\right)^2$, so the whole equation is built from just one repeated ingredient, $e^x$.
Let $u = e^x$; the equation becomes an ordinary quadratic in $u$: $$ u^2 - 5u + 4 = 0 $$
Factorise it: $(u - 1)(u - 4) = 0$, so $u = 1$ or $u = 4$.
Now undo the substitution by replacing $u$ with $e^x$ and solving each piece: $$ e^x = 1 \implies x = 0, \qquad e^x = 4 \implies x = \ln 4 $$
The exact solutions are $x = 0$ and $x = \ln 4$ (about $1.39$).

Exam technique

The quadratic formula and the discriminant $\Delta = b^2 - 4ac$ are given in the formula booklet under Topic 2, and the exponent and logarithm laws sit under Topic 1.
What is not in the booklet is the move of spotting a hidden quadratic and choosing the substitution $u = e^x$, so that recognition is something you must carry in your head.
The tell-tale sign is a term whose exponent is double another, such as $e^{2x}$ beside $e^x$, or $x^4$ beside $x^2$.

Reject Solutions That Cannot Really Exist

Substitution can hand you a value that solves the quadratic but breaks the original equation, so every candidate must be checked.
Because $u = e^x$ and an exponential is always positive, any negative value of $u$ must be thrown away.
The same care applies to logarithms: every argument inside a $\ln$ or $\log$ must stay positive, so a candidate that makes an argument zero or negative is rejected.

Common Mistake

Solving the disguised quadratic is only half the job; failing to reject an impossible root is one of the most common lost marks here.
For $u = e^x$, discard any $u \leq 0$ before taking logarithms, since $e^x > 0$ for every real $x$.
For a logarithmic equation, substitute each answer back and keep only those that leave every logarithm's argument strictly positive.

Example question

Solve $e^{2x} - 3e^x - 4 = 0$, giving your answer as an exact value.

Solution

Substitute $u = e^x$ (with $u > 0$): $u^2 - 3u - 4 = 0$.
Factorise: $(u - 4)(u + 1) = 0$, so $u = 4$ or $u = -1$.
Reject $u = -1$ because $e^x > 0$ can never be negative.
From $e^x = 4$, take logarithms: $x = \ln 4$.

Example question

Solve $\ln x + \ln(x - 3) = \ln 4$.

Solution

Combine the logarithms: $\ln\big(x(x - 3)\big) = \ln 4$.
Equate the arguments: $x^2 - 3x = 4$, so $x^2 - 3x - 4 = 0$.
Factorise: $(x - 4)(x + 1) = 0$, giving $x = 4$ or $x = -1$.
The domain needs $x > 3$ so that both $\ln x$ and $\ln(x - 3)$ are defined, so reject $x = -1$.
The only solution is $x = 4$.

Graphical Methods: When No Algebra Can Reach The Answer

Some equations have no analytic route at all, because the unknown is locked inside functions that will not combine.
The classic example is $e^x = \sin x$: no rearrangement frees $x$, so algebra simply cannot finish the job.
Here you fall back on Picture 1: draw $y = e^x$ and $y = \sin x$ on the same axes and read off where they cross.

Example

Sketch $y = e^x$, a curve that stays positive and climbs steeply for $x > 0$ while shrinking towards zero for large negative $x$.
Sketch $y = \sin x$, the familiar wave bouncing between $-1$ and $1$.
For $x \geq 0$ the exponential is already at least $1$ while the sine never exceeds $1$, so they cannot meet there.
For negative $x$ the tiny positive tail of $e^x$ slices through the sine wave again and again, giving infinitely many solutions.
Reading the first crossing to the left of the origin gives $x \approx -3.18$, and each earlier hump of the wave supplies another solution further left.

Note

To read intersections reliably, always state which two curves you drew and roughly where they meet before quoting a number.
An equation like this has no last word from algebra, so the graph is not a shortcut, it is the method.

Common Mistake

A graph read by eye or by a low-resolution calculator screen gives only approximate solutions, so quote them to three significant figures, not as if they were exact.
A calculator plot can hide a root: two curves that seem to just touch may cross twice, and a near-vertical section can conceal a solution.
Zoom in near each suspected crossing, and where you expect a whole family of solutions, say so rather than reporting just the one you can see.

Using Technology Is A Legitimate Method, Not A Last Resort

On Paper 2 a graphing calculator is not only allowed, it is often the intended tool, and "using technology" is a full method that earns full marks.
Your GDC offers three tools that all rest on the two pictures from earlier.
1. An equation solver that returns the roots of $f(x) = 0$ directly.
2. A zero finder on a graph that marks where a curve cuts the $x$-axis.
3. An intersection finder that reports where two curves cross.

Example

Take $x^4 + 5x - 6 = 0$: no factorising method in the SL toolkit unlocks a quartic like this by hand, so it is a technology problem.
Enter $y = x^4 + 5x - 6$ and ask the GDC for its zeros, or hand $x^4 + 5x - 6 = 0$ to the equation solver.
The GDC returns two real solutions, $x = -2$ and $x = 1$.
Here they happen to be whole numbers, which you can confirm by substitution: $(-2)^4 + 5(-2) - 6 = 16 - 10 - 6 = 0$ and $1 + 5 - 6 = 0$.
A quick check like this guards against mis-keying the equation, and it costs seconds.

Exam technique

On Paper 2, always write down the equation you are solving, or sketch the graph you are reading, before you quote the calculator's answer.
A bare number with no supporting line can score zero even when it is correct, because the examiner sees no method.
A good habit reads: "Solving $x^4 + 5x - 6 = 0$ using the GDC gives $x = -2$ or $x = 1$."

Example question

Use your GDC to solve $x^3 - 3x + 1 = 0$, giving each solution to three significant figures.

Solution

State the method: solve $x^3 - 3x + 1 = 0$ using the GDC equation solver or the zeros of $y = x^3 - 3x + 1$.
The curve crosses the $x$-axis three times.
The solutions are $x \approx -1.88$, $x \approx 0.347$, $x \approx 1.53$.

Choosing Your Method: Read The Equation First

Before touching algebra or a calculator, glance at the equation and ask whether a clean method exists.

Question to ask	Analytic (exact)	Technology (approximate)
Can I factorise or spot a hidden quadratic?	Yes, use it	Not needed
Can I isolate the unknown and undo it?	Yes, use logs or roots	Not needed
Is the unknown stuck inside mixed functions?	No clean route	Use GDC or a graph
Which paper am I on?	Paper 1 (no calculator)	Paper 2 (GDC allowed)
What form is the answer?	Exact: fraction, surd, ln	Decimal, 3 s.f.

Tip

If the equation is a polynomial you can factorise, or a hidden quadratic, go analytic for an exact answer.
If the unknown appears both inside and outside a function that will not combine, reach for technology.
On Paper 1 you must have an analytic route, so if you cannot find one, re-read the question for a substitution or identity you missed.

Real Situations: Where These Equations Come From

Models of the real world turn a question into an equation, and the same two toolkits solve it.
Sometimes the model rearranges cleanly, so an analytic method wins.
Sometimes the model mixes growth and straight-line change, so only technology can finish.

Example

A cup of coffee cools so that its temperature after $t$ minutes is $T = 20 + 75e^{-0.08t}$ degrees Celsius.
Asking when it reaches $50^\circ\text{C}$ means solving $20 + 75e^{-0.08t} = 50$.
This isolates cleanly, so it is an analytic problem.

Note

Two colonies are modelled by $P_A = 100 + 15t$ and $P_B = 20 \times 2^{0.5t}$, where $t$ is in years.
They are equal when $100 + 15t = 20 \times 2^{0.5t}$, an equation with $t$ both outside and inside a power, so no analytic route exists.
This is exactly the moment to state the equation and hand it to the GDC.

Example question

The temperature of a cooling drink is $T = 20 + 75e^{-0.08t}$ degrees Celsius after $t$ minutes. Find, analytically, the time when $T = 50$.

Solution

Set up the equation: $20 + 75e^{-0.08t} = 50$.
Isolate the exponential: $75e^{-0.08t} = 30$, so $e^{-0.08t} = 0.4$.
Take natural logarithms: $-0.08t = \ln 0.4$.
Solve: $t = \dfrac{\ln 0.4}{-0.08} \approx 11.5$ minutes.
So the drink reaches $50^\circ\text{C}$ after about $11.5$ minutes.

Example question

Colony A has size $P_A = 100 + 15t$ and colony B has size $P_B = 20 \times 2^{0.5t}$, with $t$ in years. Find when the two colonies are equal in size.

Solution

Set the models equal: $100 + 15t = 20 \times 2^{0.5t}$.
The unknown $t$ is both outside and inside a power, so there is no analytic method: solve with the GDC.
Find the intersection of $y = 100 + 15t$ and $y = 20 \times 2^{0.5t}$.
The curves cross at $t \approx 6.64$ years.

Active recall

Explain how solving $g(x) = h(x)$ becomes the task of finding where two graphs cross.
What substitution turns $e^{2x} - 5e^x + 4 = 0$ into a quadratic, and why is any negative value of that substitution rejected?
Why can $e^x = \sin x$ not be solved analytically, and what does its graph tell you about the number of solutions?
On Paper 2, what must you write down before quoting a solution obtained from your GDC?
Give one clue in an equation that tells you an analytic method exists, and one that tells you to use technology instead.

Why This Exists: Not Every Equation Rearranges To x = Something

Up to now, almost every equation you have solved could be rearranged step by step until $x$ sat alone on one side.
A linear equation, a factorised quadratic, or a simple exponential all give up their answer to pure algebra.
But most equations that describe the real world are not that polite: $e^x = \sin x$ cannot be rearranged to $x = \dots$ by any amount of algebra, because $x$ is trapped inside two different kinds of function at once.
So mathematicians keep two toolkits: an analytic one for equations that a method can crack, and a graphical or technology one for equations that no method can.
This subtopic is about owning both toolkits and, just as importantly, knowing on sight which one a given equation needs.

Analogy

Think of an equation like $g(x) = h(x)$ as a question: at what input do these two machines produce the same output?
Drawn as two curves, the answer is simply where they cross, so solving an equation and finding an intersections of graphs are the same act seen two ways.

Every Solution Is Either A Crossing Or A Zero

There are two pictures behind everything in this subtopic, and both turn an equation into something you can see.
Picture 1: A crossing (two curves)
1. To solve $g(x) = h(x)$, draw $y = g(x)$ and $y = h(x)$ on the same axes.
2. Each point where the curves meet has one $x$-value, and that $x$-value is a solution of the equation.
Picture 2: A zero (one curve)
1. Move everything to one side so the equation reads $f(x) = 0$, then draw the single curve $y = f(x)$.
2. Every place the curve cuts the $x$-axis is a solution, because there $y = f(x) = 0$.
The two pictures are interchangeable: $g(x) = h(x)$ becomes $g(x) - h(x) = 0$, which swaps a crossing for a zero of $f(x) = g(x) - h(x)$.

Tip

Whether you hunt for a crossing or a zero, you are looking for the same $x$-values; choose whichever picture is easier to draw.

Note

An analytic method finds these $x$-values exactly, as whole numbers, fractions, surds, or logarithms.
A graphical or technology method finds them as decimals read off a picture, usually to three significant figures.
Both are finding the same solutions; they differ only in whether the answer comes out exact or approximate.

Analytic Methods: Use Them Whenever A Method Exists

An analytic solution reaches the answer by exact algebra alone, and it is always preferred when a method is available, because the answer is exact.
Three moves cover almost every analytically solvable equation in the course.
Move 1: Factorise, then use the zero product property
1. If you can write $f(x) = 0$ as a product of factors, then a product is zero only when a factor is zero, which you meet fully when factorising and the quadratic formula.
Move 2: Rearrange and isolate, then undo
1. Get the awkward part alone on one side, then apply its inverse, for example taking $\ln$ of both sides to undo an exponential, as with exponential and logarithmic equations.
Move 3: Substitute to reveal a hidden quadratic
1. Some equations are secretly quadratics wearing a disguise, and a single substitution unmasks them.

Example

Consider $e^{2x} - 5e^x + 4 = 0$, which looks nothing like a quadratic at first glance.
Notice that $e^{2x} = \left(e^x\right)^2$, so the whole equation is built from just one repeated ingredient, $e^x$.
Let $u = e^x$; the equation becomes an ordinary quadratic in $u$: $$ u^2 - 5u + 4 = 0 $$
Factorise it: $(u - 1)(u - 4) = 0$, so $u = 1$ or $u = 4$.
Now undo the substitution by replacing $u$ with $e^x$ and solving each piece: $$ e^x = 1 \implies x = 0, \qquad e^x = 4 \implies x = \ln 4 $$
The exact solutions are $x = 0$ and $x = \ln 4$ (about $1.39$).

Exam technique

The quadratic formula and the discriminant $\Delta = b^2 - 4ac$ are given in the formula booklet under Topic 2, and the exponent and logarithm laws sit under Topic 1.
What is not in the booklet is the move of spotting a hidden quadratic and choosing the substitution $u = e^x$, so that recognition is something you must carry in your head.
The tell-tale sign is a term whose exponent is double another, such as $e^{2x}$ beside $e^x$, or $x^4$ beside $x^2$.

Reject Solutions That Cannot Really Exist

Substitution can hand you a value that solves the quadratic but breaks the original equation, so every candidate must be checked.
Because $u = e^x$ and an exponential is always positive, any negative value of $u$ must be thrown away.
The same care applies to logarithms: every argument inside a $\ln$ or $\log$ must stay positive, so a candidate that makes an argument zero or negative is rejected.

Common Mistake

Solving the disguised quadratic is only half the job; failing to reject an impossible root is one of the most common lost marks here.
For $u = e^x$, discard any $u \leq 0$ before taking logarithms, since $e^x > 0$ for every real $x$.
For a logarithmic equation, substitute each answer back and keep only those that leave every logarithm's argument strictly positive.

Example question

Solve $e^{2x} - 3e^x - 4 = 0$, giving your answer as an exact value.

Solution

Substitute $u = e^x$ (with $u > 0$): $u^2 - 3u - 4 = 0$.
Factorise: $(u - 4)(u + 1) = 0$, so $u = 4$ or $u = -1$.
Reject $u = -1$ because $e^x > 0$ can never be negative.
From $e^x = 4$, take logarithms: $x = \ln 4$.

Example question

Solve $\ln x + \ln(x - 3) = \ln 4$.

Solution

Combine the logarithms: $\ln\big(x(x - 3)\big) = \ln 4$.
Equate the arguments: $x^2 - 3x = 4$, so $x^2 - 3x - 4 = 0$.
Factorise: $(x - 4)(x + 1) = 0$, giving $x = 4$ or $x = -1$.
The domain needs $x > 3$ so that both $\ln x$ and $\ln(x - 3)$ are defined, so reject $x = -1$.
The only solution is $x = 4$.

Graphical Methods: When No Algebra Can Reach The Answer

Some equations have no analytic route at all, because the unknown is locked inside functions that will not combine.
The classic example is $e^x = \sin x$: no rearrangement frees $x$, so algebra simply cannot finish the job.
Here you fall back on Picture 1: draw $y = e^x$ and $y = \sin x$ on the same axes and read off where they cross.

Example

Sketch $y = e^x$, a curve that stays positive and climbs steeply for $x > 0$ while shrinking towards zero for large negative $x$.
Sketch $y = \sin x$, the familiar wave bouncing between $-1$ and $1$.
For $x \geq 0$ the exponential is already at least $1$ while the sine never exceeds $1$, so they cannot meet there.
For negative $x$ the tiny positive tail of $e^x$ slices through the sine wave again and again, giving infinitely many solutions.
Reading the first crossing to the left of the origin gives $x \approx -3.18$, and each earlier hump of the wave supplies another solution further left.

Note

To read intersections reliably, always state which two curves you drew and roughly where they meet before quoting a number.
An equation like this has no last word from algebra, so the graph is not a shortcut, it is the method.

Common Mistake

A graph read by eye or by a low-resolution calculator screen gives only approximate solutions, so quote them to three significant figures, not as if they were exact.
A calculator plot can hide a root: two curves that seem to just touch may cross twice, and a near-vertical section can conceal a solution.
Zoom in near each suspected crossing, and where you expect a whole family of solutions, say so rather than reporting just the one you can see.

Using Technology Is A Legitimate Method, Not A Last Resort

On Paper 2 a graphing calculator is not only allowed, it is often the intended tool, and "using technology" is a full method that earns full marks.
Your GDC offers three tools that all rest on the two pictures from earlier.
1. An equation solver that returns the roots of $f(x) = 0$ directly.
2. A zero finder on a graph that marks where a curve cuts the $x$-axis.
3. An intersection finder that reports where two curves cross.

Example

Take $x^4 + 5x - 6 = 0$: no factorising method in the SL toolkit unlocks a quartic like this by hand, so it is a technology problem.
Enter $y = x^4 + 5x - 6$ and ask the GDC for its zeros, or hand $x^4 + 5x - 6 = 0$ to the equation solver.
The GDC returns two real solutions, $x = -2$ and $x = 1$.
Here they happen to be whole numbers, which you can confirm by substitution: $(-2)^4 + 5(-2) - 6 = 16 - 10 - 6 = 0$ and $1 + 5 - 6 = 0$.
A quick check like this guards against mis-keying the equation, and it costs seconds.

Exam technique

On Paper 2, always write down the equation you are solving, or sketch the graph you are reading, before you quote the calculator's answer.
A bare number with no supporting line can score zero even when it is correct, because the examiner sees no method.
A good habit reads: "Solving $x^4 + 5x - 6 = 0$ using the GDC gives $x = -2$ or $x = 1$."

Example question

Use your GDC to solve $x^3 - 3x + 1 = 0$, giving each solution to three significant figures.

Solution

State the method: solve $x^3 - 3x + 1 = 0$ using the GDC equation solver or the zeros of $y = x^3 - 3x + 1$.
The curve crosses the $x$-axis three times.
The solutions are $x \approx -1.88$, $x \approx 0.347$, $x \approx 1.53$.

Choosing Your Method: Read The Equation First

Before touching algebra or a calculator, glance at the equation and ask whether a clean method exists.

Question to ask	Analytic (exact)	Technology (approximate)
Can I factorise or spot a hidden quadratic?	Yes, use it	Not needed
Can I isolate the unknown and undo it?	Yes, use logs or roots	Not needed
Is the unknown stuck inside mixed functions?	No clean route	Use GDC or a graph
Which paper am I on?	Paper 1 (no calculator)	Paper 2 (GDC allowed)
What form is the answer?	Exact: fraction, surd, ln	Decimal, 3 s.f.

Tip

If the equation is a polynomial you can factorise, or a hidden quadratic, go analytic for an exact answer.
If the unknown appears both inside and outside a function that will not combine, reach for technology.
On Paper 1 you must have an analytic route, so if you cannot find one, re-read the question for a substitution or identity you missed.

Real Situations: Where These Equations Come From

Models of the real world turn a question into an equation, and the same two toolkits solve it.
Sometimes the model rearranges cleanly, so an analytic method wins.
Sometimes the model mixes growth and straight-line change, so only technology can finish.

Example

A cup of coffee cools so that its temperature after $t$ minutes is $T = 20 + 75e^{-0.08t}$ degrees Celsius.
Asking when it reaches $50^\circ\text{C}$ means solving $20 + 75e^{-0.08t} = 50$.
This isolates cleanly, so it is an analytic problem.

Note

Two colonies are modelled by $P_A = 100 + 15t$ and $P_B = 20 \times 2^{0.5t}$, where $t$ is in years.
They are equal when $100 + 15t = 20 \times 2^{0.5t}$, an equation with $t$ both outside and inside a power, so no analytic route exists.
This is exactly the moment to state the equation and hand it to the GDC.

Example question

The temperature of a cooling drink is $T = 20 + 75e^{-0.08t}$ degrees Celsius after $t$ minutes. Find, analytically, the time when $T = 50$.

Solution

Set up the equation: $20 + 75e^{-0.08t} = 50$.
Isolate the exponential: $75e^{-0.08t} = 30$, so $e^{-0.08t} = 0.4$.
Take natural logarithms: $-0.08t = \ln 0.4$.
Solve: $t = \dfrac{\ln 0.4}{-0.08} \approx 11.5$ minutes.
So the drink reaches $50^\circ\text{C}$ after about $11.5$ minutes.

Example question

Colony A has size $P_A = 100 + 15t$ and colony B has size $P_B = 20 \times 2^{0.5t}$, with $t$ in years. Find when the two colonies are equal in size.

Solution

Set the models equal: $100 + 15t = 20 \times 2^{0.5t}$.
The unknown $t$ is both outside and inside a power, so there is no analytic method: solve with the GDC.
Find the intersection of $y = 100 + 15t$ and $y = 20 \times 2^{0.5t}$.
The curves cross at $t \approx 6.64$ years.

Active recall

Explain how solving $g(x) = h(x)$ becomes the task of finding where two graphs cross.
What substitution turns $e^{2x} - 5e^x + 4 = 0$ into a quadratic, and why is any negative value of that substitution rejected?
Why can $e^x = \sin x$ not be solved analytically, and what does its graph tell you about the number of solutions?
On Paper 2, what must you write down before quoting a solution obtained from your GDC?
Give one clue in an equation that tells you an analytic method exists, and one that tells you to use technology instead.

IB Mathematics Analysis and Approaches SL 2.10—solving Equations Graphically and Analytically Notes

IB Mathematics Analysis and Approaches SL 2.10—solving Equations Graphically and Analytically Notes

Why This Exists: Not Every Equation Rearranges To x = Something

Every Solution Is Either A Crossing Or A Zero

Analytic Methods: Use Them Whenever A Method Exists

Reject Solutions That Cannot Really Exist

Graphical Methods: When No Algebra Can Reach The Answer

Using Technology Is A Legitimate Method, Not A Last Resort

Choosing Your Method: Read The Equation First

Real Situations: Where These Equations Come From

Why This Exists: Not Every Equation Rearranges To x = Something

Every Solution Is Either A Crossing Or A Zero

Analytic Methods: Use Them Whenever A Method Exists

Reject Solutions That Cannot Really Exist

Graphical Methods: When No Algebra Can Reach The Answer

Using Technology Is A Legitimate Method, Not A Last Resort

Choosing Your Method: Read The Equation First

Real Situations: Where These Equations Come From

Number and Algebra

Functions

Geometry & Trigonometry

Statistics & Probability

Calculus

IB Mathematics Analysis and Approaches SL 2.10—solving Equations Graphically and Analytically Notes

Number and Algebra

Functions

Geometry & Trigonometry

Statistics & Probability

Calculus

IB Mathematics Analysis and Approaches SL 2.10—solving Equations Graphically and Analytically Notes

Why This Exists: Not Every Equation Rearranges To x = Something

Every Solution Is Either A Crossing Or A Zero

Analytic Methods: Use Them Whenever A Method Exists

Reject Solutions That Cannot Really Exist

Graphical Methods: When No Algebra Can Reach The Answer

Using Technology Is A Legitimate Method, Not A Last Resort

Choosing Your Method: Read The Equation First

Real Situations: Where These Equations Come From

Why This Exists: Not Every Equation Rearranges To x = Something

Every Solution Is Either A Crossing Or A Zero

Analytic Methods: Use Them Whenever A Method Exists

Reject Solutions That Cannot Really Exist

Graphical Methods: When No Algebra Can Reach The Answer

Using Technology Is A Legitimate Method, Not A Last Resort

Choosing Your Method: Read The Equation First

Real Situations: Where These Equations Come From