Find the x that makes it true: Solving an equation means finding the value(s) of x that make both sides equal.
You can do it analytically (algebra) or graphically (read the picture / GDC) — both give the same answers.
Analytic (algebra)
- Rearrange and solve exactly
- Factor, formula, logs…
- Best on Paper 1 (no GDC)
Graphical (GDC)
- Read solutions off a graph
- zeros / intersect tools
- Best on Paper 2
Check your answer: Substitute a solution back in — both sides should match.
A quick sanity check catches slips.
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Get everything on one side first: A reliable universal move: bring every term to one side so the other is 0, then find the roots (factor, quadratic formula, or GDC).
Zero on one side turns 'solve' into 'find the zeros'.
IB-style question — rearrange then solve
Solve x² + 2 = 5x.
Step by step
- Move everything to one side.
- Use the quadratic formula (it doesn't factor nicely).
Final answer
x = (5 ± √17)/2 ≈ 4.56 or 0.438.
Don't divide by x: Dividing by x can lose the solution x = 0.
Move terms across and factor instead.
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Solutions of f(x) = 0 are the x-intercepts: Graph y = f(x): the solutions of f(x) = 0 are exactly the x-intercepts (zeros) — where the curve crosses the x-axis.
On Paper 2, use the GDC's zero tool.
Watch it build: graph y = x² − 5x + 6, and its x-intercepts (2, 0) and (3, 0) are exactly the solutions of x² − 5x + 6 = 0.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
GDC walkthrough
Step through the exact calculator keystrokes, screen by screen, in study mode.
Equate, or find where the graphs meet: To solve f(x) = g(x): algebraically, set them equal and rearrange to = 0; graphically, the solutions are the x-coordinates of the intersection points of y = f(x) and y = g(x).
IB-style question — two methods
Solve 2ˣ = x + 3 (you may use a GDC).
Step by step
- This mixes an exponential and a line — no neat algebra.
- Graph y = 2ˣ and y = x + 3 and use intersect.
Final answer
x ≈ −2.9 or x ≈ 2.44 (from the GDC).
When algebra runs out, graph it: Equations mixing different function types (exponential and linear, etc.) usually can't be solved by algebra — the GDC graph/intersect is the intended method.