2.10 Solving equations graphically and analytically (IB Maths: Analysis and Approaches SL): Revision notes

Section 1

Solving analytically: disguised quadratics

Many exponential equations are quadratics in disguise. Look for a term that is the square of another:

$e^{2x} = (e^{x})^{2}$ , so let $u = e^{x}$ .
$9^{x} = (3^{x})^{2}$ and $4^{x} = (2^{x})^{2}$ .

$e^{2x} - 5e^{x} + 4 = 0 \Rightarrow u^{2} - 5u + 4 = 0 \Rightarrow u = 1$ or $u = 4$ , so $x = \ln 1 = 0$ or $x = \ln 4$ .

Always substitute back and check each value of $u$ : since $e^{x} > 0$ and $a^{x} > 0$ , any negative or zero value of $u$ gives no solution.

Key termssubstitutionanalytic solution

Common mistake

Stopping at $u = 1$ or $u = 4$ . The question asks for $x$ , so solve $e^{x} = 1$ and $e^{x} = 4$ .

Common mistake

Keeping $e^{x} = -3$ . An exponential is always positive, so reject it and give a reason.

Section 2

Solving graphically with a GDC

Some equations, such as $e^{x} = x + 2$ , $e^{x} = \sin x$ or $x^{4} + 5x - 6 = 0$ , have no appropriate analytic method. Use technology:

Graph $y =$ left side and $y =$ right side, and find the $x$ -coordinates of the intersections; or
Rearrange to $h(x) = 0$ and find the zeros of $h$ .

Give answers to 3 significant figures unless told otherwise, and give every solution: check the window is wide enough to see them all.

Key termsintersectionzero

Common mistake

Giving the $y$ -coordinate of an intersection instead of the $x$ -coordinate. For $e^{x} = x + 2$ the solution is $x \approx 1.15$ , not $3.15$ .

Exam tip

Before using the GDC, reason about how many solutions to expect: $e^{x} < x + 2$ at $x = 0$ but $e^{x} > x + 2$ far to each side, so there are two.

Section 3

Inequalities from graphs

Once the intersections are known, an inequality is solved by deciding which graph is above the other in each region. For $e^{x} > x + 2$ with intersections at $x \approx -1.84$ and $x \approx 1.15$ , the exponential is above the line outside these values: $x < -1.84$ or $x > 1.15$ .

Analytically, work with the substitution: $e^{2x} < 6e^{x} - 8 \Leftrightarrow 2 < e^{x} < 4 \Leftrightarrow \ln 2 < x < \ln 4$ .

Key termscritical values

Section 4

Equations with a parameter

If a disguised quadratic has a parameter, such as $u^{2} - 6u + (8 - k) = 0$ with $u = e^{x}$ , count solutions for $x$ carefully:

each positive root $u$ gives exactly one $x$ ;
zero or negative roots give none.

So 'exactly one solution' can come from a repeated positive root ( $\Delta = 0$ ), or from one positive and one non-positive root. Check the sign of the roots, not just the discriminant.

Key termsparameter

Exam tip

For $u^{2} + bu + c = 0$ , the product of the roots is $c$ and the sum is $-b$ . If $c > 0$ and $-b > 0$ , both real roots are positive.

Section 5

Equations in real-life contexts

Models often lead to equations such as $40e^{0.02t} = 65 - 0.5t$ (when do two populations become equal?). Use a GDC, then interpret:

Convert $t$ to a real time: if $t$ is years after 1 January 2020, then $t = 17.2$ falls during 2037.
State units and sensible accuracy (e.g. 56.4 thousand people).
Comment on validity: a linear model $65 - 0.5t$ becomes negative for $t > 130$ , which is impossible for a population.

Key termsvalidity

Common mistake

Adding $t$ to the year incorrectly: $t = 17.2$ years after the start of 2020 is early 2037, not 2017 or 2038.

Must know

Spot hidden quadratics: $e^{2x} = (e^{x})^{2}$ , $9^{x} = (3^{x})^{2}$ ; substitute, solve, substitute back.
Reject values of $u$ that make $e^{x}$ or $a^{x}$ zero or negative, with a reason.
With a GDC: use intersections or zeros; find every solution; 3 significant figures.
Inequalities: find intersections, then decide which graph is above.
In context, interpret solutions in real units and comment on the model's validity.

That's the notes covered.

Carry on to the next subtopic.