1The idea in one paragraph

There are two ways to solve an equation. Analytically means by algebra: a chain of steps, each one undoing something, ending with an exact answer such as x = ln 5. Graphically means by looking at graphs: the solutions of f(x) = g(x) are the x-coordinates where y = f(x) and y = g(x) cross, which technology finds to any accuracy you need. Algebra gives exact answers but only works for equations with the right shape. Graphs work for every equation but give decimals, and they show you something algebra hides: how many solutions there are. The skill is recognising which route a question wants, and using both together when that helps: the algebra for the answer, the graph to check you have them all.

2Two routes, and when each is expected

The words of the question decide. "Find the exact value", "show that" and "hence" on Paper 1 all require algebra. On Paper 2, "solve" with no further instruction lets you use the GDC, and for many equations it is the only way.

3Equations that are quadratics in disguise

Many Paper 1 equations are quadratics wearing a costume. Spot the pattern "something squared, the same something, a constant", replace the something with a single letter, and solve the quadratic.

Exponentials. Since e²ˣ = (eˣ)², the equation e²ˣ − 7eˣ + 10 = 0 is a quadratic in eˣ.

Figure 1 shows the graph of y = e²ˣ − 7eˣ + 10 crossing the x-axis at exactly those two values, and nowhere else.

Rejecting an impossible root. The quadratic can produce a value of u that the original letter cannot take. In e²ˣ − eˣ − 6 = 0:

Writing "no solution since eˣ > 0" is not decoration: it is the reasoning mark.

Other bases. In 4ˣ − 6 × 2ˣ + 8 = 0, notice 4ˣ = (2²)ˣ = (2ˣ)². With u = 2ˣ: u² − 6u + 8 = 0, so u = 2 or u = 4, so 2ˣ = 2 or 2ˣ = 4, giving x = 1 or x = 2.

Even powers. x⁴ − 5x² + 4 = 0 is a quadratic in x². With u = x²: (u − 1)(u − 4) = 0, so x² = 1 or x² = 4, giving x = ±1, ±2: four solutions. If one of the values of u had been negative, x² = negative would have had no real solutions.

Fractions. x + 6/x = 5 becomes a quadratic when you multiply through by x (allowed, because x = 0 is not in the domain): x² + 6 = 5x, so x² − 5x + 6 = 0, giving x = 2 or x = 3.

4Exponential and logarithmic equations

The unknown in the exponent: take logs of both sides.

With the unknown on both sides, take logs and collect the x terms:

The unknown inside logarithms: combine the logs, then exponentiate. Use the laws of 1.7 to make a single logarithm, then undo it.

Now check the domain. The original equation contains log₂ x and log₂(x − 2), so it needs x > 0 and x > 2. The value x = −2 makes both logarithms undefined, so it is rejected. x = 4 is the only solution.

Every solution must be checked against the original equation's domain. Combining logarithms and squaring can create values that were never solutions.

The reason false roots appear: log₂(x(x − 2)) is defined for x = −2 (the product is 8), even though log₂ x and log₂(x − 2) separately are not. Combining the logarithms widened the domain, and the quadratic found a solution in the wider one.

5Solving with technology

Most equations met in real modelling have no algebraic method at all. There is no way to rearrange ln x = 3 − x to get x on its own; the same goes for eˣ = x + 2, 2 sin x = x, or a cubic like x³ − 4x + 1 = 0 that does not factorise. For these, Paper 2 expects the GDC, in one of two ways from 2.4:

• Intersect. Graph y = f(x) and y = g(x) and use the intersect tool at each crossing.
• Zeros. Rearrange to f(x) − g(x) = 0, graph y = f(x) − g(x), and use the zero tool at each crossing of the x-axis.

For ln x = 3 − x, Figure 2 shows the two graphs crossing once, at (2.21, 0.792). The solution is x = 2.21 (3 s.f.). Since ln x is increasing and 3 − x is decreasing, they can only cross once, so this is the only solution.

For x³ − 4x + 1 = 0, graph y = x³ − 4x + 1 and use the zero tool three times: x = −2.11, 0.254, 1.86. A cubic can have up to three real roots, so after finding one, keep looking.

For eˣ = x + 2, the intersections are at x = −1.84 and x = 1.15: the left-hand one is easy to miss, because the exponential there is nearly flat against the x-axis and the default window may not show the crossing clearly.

Trigonometric equations: set the mode. Figure 3 shows 2 sin x = x solved with the calculator in radians: three solutions, x = −1.90, 0, 1.90. In degree mode the calculator would be solving a different equation, since sin 1.90° is not sin 1.90. Unless the question gives angles in degrees, use radians.

Four habits make GDC answers score.

1. Choose a window that shows every solution. Zoom out first, count the crossings, then zoom in. If the question gives a domain, such as −π ≤ x ≤ π, set the window to exactly that domain.
2. Use the tool once per solution. Each use of zero or intersect gives one answer.
3. Write the method. "Using GDC to intersect y = ln x and y = 3 − x" earns the method mark even if a value is miscopied.
4. 3 significant figures, but store the full value if it is used again.

6How many solutions? Reading it from a graph

A graph answers a question algebra finds awkward: how many solutions does an equation have? The solutions of f(x) = k are where the horizontal line y = k meets the graph of f. Slide the line up and down and count.

Take f(x) = x³ − 3x. The GDC (or calculus later) gives a local maximum at (−1, 2) and a local minimum at (1, −2). Figure 4 shows three lines:

• For −2 < k < 2 the line cuts the curve three times: three solutions.
• For k = 2 or k = −2 it touches at a turning point and crosses once elsewhere: two solutions.
• For k > 2 or k < −2 it meets the curve once: one solution.

The turning values are the boundaries every time. This is the same reasoning as the discriminant in 2.7, which counted the meetings of a parabola with a line, now for any curve.

Counting also works with two curves. ln x = 3 − x has one solution because an increasing and a decreasing graph can cross at most once. An argument like that, in words, is what a question means by "explain why the equation has exactly one solution".

7Equations from real situations

The guide asks you to solve equations that arise from contexts: decay, population growth, compound interest, projectiles, braking distances. The pattern is always the same: translate the question into an equation, solve it by the route it allows, then translate the answer back, with units and a check that it makes sense.

Braking distance. A car's stopping distance in metres is modelled by d(v) = 0.2v + 0.005v², where v is its speed in km/h. What is the greatest speed at which it can stop within 40 m?

Speed is positive, so the greatest speed is 71.7 km/h. The negative root is a solution of the equation but not of the problem.

Two populations. Town A has 12 000 people and grows at a continuous 2.5% a year: A(t) = 12 000e^(0.025t). Town B has 15 000 people and grows by 200 people a year: B(t) = 15 000 + 200t. When does A overtake B?

The equation 12 000e^(0.025t) = 15 000 + 200t mixes an exponential with a linear term, so there is no algebraic method. Intersect the two graphs on the GDC (Figure 5): t = 17.2, when both towns have about 18 400 people. Town A overtakes town B after about 17.2 years. A percentage growth always beats a fixed yearly increase eventually, however far behind it starts.

Two last checks in any context. Is the answer inside the domain where the model makes sense (positive times, realistic sizes)? And did the question ask for an x-value, a point, or a quantity in the context's units?

8Choosing a method

Figure 6 puts the decisions of this page in one place.

Before anything else, rearrange: either one side zero or the form f(x) = g(x). Then ask whether the equation has a shape you recognise (a quadratic, a quadratic in eˣ or x², a single exponential or logarithm). If it does and the answer must be exact, use algebra, check every root against the domain, and on Paper 2 glance at the graph to be sure you have them all. If the equation mixes kinds of function, eˣ with x, ln x with x, sin x with x, it has no algebraic method on this syllabus: graph it, and find every crossing.