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Mathematics AI · HL · Unit 1: Number and Algebra

Unit 1Topic 1.8Key Concepts62 min read

Solving equations with technology

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Equations, solutions and the role of technology

An equation states that two expressions are equal. A solution is a value, or ordered set of values, that makes the equation true. Solving an equation means finding the complete solution set within the stated domain—not merely accepting the first decimal shown by a calculator.

In AI HL, technology is a mathematical tool rather than a substitute for reasoning. A GDC can solve, graph and tabulate, but the student must enter the correct model, choose the variables and domain, identify every relevant solution, and explain what the result means.

The core syllabus requirement is to use technology to solve systems of linear equations in up to three variables and polynomial equations. No single calculator method is prescribed, so graphing, equation solvers and polynomial/system tools are all acceptable when used correctly.

A calculator output is initially a candidate solution. It becomes a validated solution only after it satisfies the original equation, all domain restrictions and any contextual conditions such as t≥0t\ge 0 or an integer number of items.

Key terms

Equation
A statement that two mathematical expressions have equal value.
Solution
A value or ordered set of values that makes the original equation or system true.
Solution set
All valid solutions in the specified domain.
Candidate solution
A possible answer returned or suggested by a numerical method before validation.
Numerical solution
A decimal approximation obtained using computation, graphing, a table or an iterative solver.
Domain
The set of input values allowed by the mathematics or context.

Worked examples

Check a proposed solution

Is x=4x=4 a solution of 2x+3=112x+3=11?

  1. 1Substitute x=4x=4 into the left side: 2(4)+3=112(4)+3=11.
  2. 2The left side equals the right side.
  3. 3Therefore the proposed value satisfies the original equation.

Answer: Yes. The solution is x=4x=4.

Reject a contextually invalid candidate

A model for the side length of a square produces x=−6x=-6 and x=6x=6. Which solution is valid?

  1. 1Both values may solve the algebraic equation.
  2. 2A physical length cannot be negative.
  3. 3Apply the contextual domain x>0x\gt 0.

Answer: The valid contextual solution is x=6x=6.

Zeros, roots and intersections: equivalent views

For one-variable equations, the words root and zero refer to a value x=rx=r for which h(r)=0h(r)=0. On the graph y=h(x)y=h(x), the point (r,0)(r,0) is an xx-intercept.

Any equation f(x)=g(x)f(x)=g(x) can be rearranged to h(x)=f(x)−g(x)=0h(x)=f(x)-g(x)=0. Therefore the same solution can be found either where y=f(x)y=f(x) and y=g(x)y=g(x) intersect, or where y=h(x)y=h(x) meets the xx-axis.

Five-step GDC workflow: define, choose a method, frame the domain and window, solve, then check.Technology finds candidates; mathematical checking determines whether they are valid solutions.

The zero form is often convenient when a GDC has a root or polynomial tool. The intersection form is often more natural when the two sides represent different real-world quantities, such as revenue and cost.

Key terms

Root or zero
An input r such that h(r)=0h(r)=0.
x-intercept
A graph point r,0r,0, corresponding to a real root r.
Intersection
A point at which two graphs have the same x- and y-coordinates.
Equivalent equation
A reformulation with the same solution set, subject to the same restrictions.

Worked examples

One equation, two graph methods

Solve x2−x−6=0x^2-x-6=0 using zero and intersection language.

  1. 1Zero method: graph h(x)=x2−x−6h(x)=x^2-x-6 and find its xx-intercepts.
  2. 2Intersection method: rewrite as x2=x+6x^2=x+6, then graph y1=x2y_1=x^2 and y2=x+6y_2=x+6.
  3. 3Both methods produce the same xx-coordinates: −2-2 and 33.
  4. 4Substitution confirms both values satisfy the original equation.

Answer: x=−2x=-2 or x=3x=3.

Build a zero function

Rewrite 5(0.8)x=2x+15(0.8)^x=2x+1 as a zero problem.

  1. 1Move one side to the other without changing the equality.
  2. 2Define h(x)=5(0.8)x−(2x+1)h(x)=5(0.8)^x-(2x+1).
  3. 3The original solutions are exactly the zeros of hh.

Answer: Solve h(x)=5(0.8)x−2x−1=0h(x)=5(0.8)^x-2x-1=0.

Rules and formulas

Root definition

h(r)=0h(r)=0

Intersection-to-zero conversion

f(x)=g(x)  ⟺  f(x)−g(x)=0f(x)=g(x)\iff f(x)-g(x)=0

A reliable numerical-solving workflow

A numerical solver searches for decimal values that make an equation approximately true. Different calculators may use different internal algorithms, but the student workflow is stable: define the equation, restrict the search, choose a method, calculate candidates, and check them.

Start by forming either h(x)=0h(x)=0 or f(x)=g(x)f(x)=g(x). Estimate where solutions may lie using algebra, context, a graph or a table. A solver may need an initial guess or lower and upper bounds; poor choices can lead to a different root or no result.

After solving, keep more digits than you plan to report. Substitute the unrounded value into the original equation, then round only the final result according to the question.

A numerical method can miss roots when the search interval is too narrow, when the graph only touches the axis, or when multiple roots lie near one another. A complete solution needs deliberate searching, not one button press.

Key terms

Initial guess
A starting value supplied to a numerical solver.
Search interval
The range within which technology looks for a solution.
Iteration
Repeated computational improvement of an approximation.
Convergence
The process by which successive approximations approach a stable solution.
Residual
The numerical difference between the two sides after substituting a candidate.

Worked examples

Solve an equation with no convenient elementary exact form

Solve x3−x−1=0x^3-x-1=0.

  1. 1Define h(x)=x3−x−1h(x)=x^3-x-1.
  2. 2A table gives h(1)=−1h(1)=-1 and h(2)=5h(2)=5, so search on 1<x<21\lt x\lt2.
  3. 3Use a zero or numerical-solver command in that interval.
  4. 4The calculator gives x≈1.324717957x\approx1.324717957.
  5. 5Substitute the stored value to confirm the residual is essentially zero.

Answer: x≈1.325x\approx1.325 to three decimal places.

Use different guesses to find different roots

A graph suggests that x3−4x−1=0x^3-4x-1=0 has three real roots. How should a solver be used?

  1. 1Use a graph or table to identify three separate neighborhoods.
  2. 2Search near −1.9-1.9, −0.3-0.3 and 2.12.1, or bracket each root separately.
  3. 3The three results are approximately −1.860806-1.860806, −0.254102-0.254102 and 2.1149082.114908.
  4. 4Check each candidate in the original cubic and report all three.

Answer: x≈−1.861,−0.254,2.115x\approx-1.861,-0.254,2.115.

Rules and formulas

Absolute residual

R=∣LHS−RHS∣R=\left|\text{LHS}-\text{RHS}\right|

Using graphs to find zeros and roots

To solve h(x)=0h(x)=0 graphically, enter Y1=h(x)Y_1=h(x), display a suitable window, and use the GDC zero/root command. The cursor or bracket should enclose the intended root. Record the xx-coordinate, not merely the screen point.

A graph crossing the axis usually indicates a sign change. A repeated root may only touch the axis and turn around, so a sign-change table can miss it. Zooming near a suspected tangency and using a polynomial solver provide useful confirmation.

A graph is an approximation. Pixel resolution and an unsuitable scale can make close roots appear as one root or a shallow crossing appear to miss the axis. Use a numerical command for the final decimal.

Key terms

Crossing root
A root at which the graph passes from one side of the x-axis to the other.
Repeated root
A root whose factor occurs more than once; an even-multiplicity root often touches the axis.
Multiplicity
The number of times a root's factor is repeated.
Root command
A GDC function that numerically locates an x-intercept in a selected interval.

Worked examples

Find all roots of a cubic from a graph

Use technology to solve x3−4x−1=0x^3-4x-1=0.

  1. 1Graph Y1=x3−4x−1Y_1=x^3-4x-1 on a window such as −3≤x≤3-3\le x\le3.
  2. 2The graph shows three axis crossings.
  3. 3Use the zero command separately around each crossing.
  4. 4Store the results and substitute them into the polynomial.

Answer: x≈−1.861,−0.254,2.115x\approx-1.861,-0.254,2.115.

Detect a touching root

Solve (x−2)2(x+1)=0(x-2)^2(x+1)=0 from its graph.

  1. 1Graph Y1=(x−2)2(x+1)Y_1=(x-2)^2(x+1).
  2. 2The graph crosses at x=−1x=-1.
  3. 3At x=2x=2, the graph touches the axis and turns because the factor ((x-2)) is squared.
  4. 4Confirm both roots with a polynomial tool or substitution.

Answer: The roots are x=−1x=-1 and x=2x=2; x=2x=2 has multiplicity 22.

Rules and formulas

Polynomial factor and root

(x−r)m is a factor⟹r is a root of multiplicity m(x-r)^m\text{ is a factor}\Longrightarrow r\text{ is a root of multiplicity }m

Finding intersections of functions

To solve f(x)=g(x)f(x)=g(x), enter Y1=f(x)Y_1=f(x) and Y2=g(x)Y_2=g(x), display both graphs, and use an intersection command. The calculator returns a point (x,y)(x,y); the equation's solution is normally the xx-coordinate.

Intersection methods are especially useful when each side has a contextual meaning—for example, revenue and cost, supply and demand, or two competing population models. The full point may then be meaningful.

As with roots, there may be several intersections. Pan, zoom, change the domain, and use a table of differences to make sure the full relevant domain has been searched.

Key terms

Intersection solution
The x-coordinate where two functions have equal output.
Break-even point
An input at which revenue equals cost.
Difference function
The function h(x)=f(x)−g(x)h(x)=f(x)-g(x), whose zeros are intersections of ff and gg.

Worked examples

Linear-exponential intersection

Solve 0.5x+2=8(0.7)x0.5x+2=8(0.7)^x for x≥0x\ge0.

  1. 1Graph Y1=0.5x+2Y_1=0.5x+2 and Y2=8(0.7)xY_2=8(0.7)^x.
  2. 2Use the domain x≥0x\ge0 and choose a window that displays both graphs.
  3. 3Select the intersection command near the visible crossing.
  4. 4The GDC gives approximately (2.517831,3.258915)(2.517831,3.258915).
  5. 5Substitute the stored x-value into both functions to verify equal outputs.

Answer: x≈2.518x\approx2.518. The common output is approximately 3.2593.259.

Break-even intersection

Revenue is R(x)=1200xR(x)=1200x and cost is C(x)=30000+15x1.5C(x)=30000+15x^{1.5}, where x≥0x\ge0 is the number of units. Estimate all continuous-model break-even inputs and identify the profitable whole-unit range.

  1. 1Solve 1200x=30000+15x321200x=30000+15x^{\frac{3}{2}} on x≥0x\ge0. A wider graph window reveals two intersections, near 26.727189326.7271893 and 6349.70315286349.7031528.
  2. 2Profit P(x)=1200x−30000−15x32P(x)=1200x-30000-15x^{\frac{3}{2}} has P′(x)=1200−22.5xP'(x)=1200-22.5\sqrt{x}, so it increases then decreases, allowing no further roots on the nonnegative domain.
  3. 3Check profit signs around both intersections. Profit is positive between them and negative outside.
  4. 4Since units are integers, the profitable range is 27≤x≤634927\le x\le6349. Exact zero profit does not occur at an integer in this model.

Answer: Continuous break-even inputs are approximately 26.727226.7272 and 6349.70326349.7032. The first profitable whole-unit output is 2727; profitable integer outputs run through 63496349.

Rules and formulas

Break-even equation

R(x)=C(x)R(x)=C(x)

Difference function for intersections

h(x)=f(x)−g(x)h(x)=f(x)-g(x)

Using tables to locate and bracket solutions

A value table is useful when a graph scale hides detail or a numerical solver needs bounds. Evaluate the zero function at regularly spaced inputs and look for values close to zero or a change of sign.

If hh is continuous and h(a)h(b)<0h(a)h(b)\lt0, then at least one root lies between aa and bb. This sign-change test locates a bracket but does not prove there is exactly one root.

Reduce the table step size to narrow the interval. A coarse table can skip two nearby crossings, and it can miss a touching root because no sign change occurs.

For an intersection f(x)=g(x)f(x)=g(x), tabulate both functions or tabulate h(x)=f(x)−g(x)h(x)=f(x)-g(x). A sign change in hh brackets an intersection.

Key terms

Table step
The increment between consecutive x-values in a numerical table.
Bracket
An interval chosen to contain a root.
Sign change
A switch between positive and negative function values across an interval.
Continuous function
A function with no break over the interval under consideration.

Worked examples

Bracket a cubic root

Use a table to locate the real root of x3−x−1=0x^3-x-1=0.

  1. 1Define h(x)=x3−x−1h(x)=x^3-x-1.
  2. 2The table shows h(1)=−1h(1)=-1 and h(2)=5h(2)=5, so a root lies in (1,2)(1,2).
  3. 3With step 0.10.1, h(1.3)≈−0.103h(1.3)\approx-0.103 and h(1.4)≈0.344h(1.4)\approx0.344.
  4. 4Use 1.3 and 1.4 as solver bounds, or refine the table further.
  5. 5The numerical root is 1.324717957…1.324717957\ldots.

Answer: x≈1.325x\approx1.325.

Recognize the limitation of a sign-change table

Why might a table fail to reveal the root x=2x=2 of h(x)=(x−2)2h(x)=(x-2)^2?

  1. 1For all x≠2x\ne2, h(x)>0h(x)\gt0.
  2. 2Values on both sides of 2 have the same sign.
  3. 3If the table does not include x=2x=2 exactly, it shows small positive values but no sign change.

Answer: A sign-change search can miss a touching root; inspect minima or use a polynomial/root tool.

Rules and formulas

Sign-change bracket

h(a)h(b)<0⟹at least one root in (a,b) if h is continuoush(a)h(b)\lt0\Longrightarrow\text{at least one root in }(a,b)\text{ if }h\text{ is continuous}

Choosing suitable domains and graph windows

A graph window is not the mathematical domain. It is only the portion of the coordinate plane displayed. If a solution lies outside the window, the graph can falsely appear to have no solution.

Choose the x-range from algebra or context. Time usually begins at zero; a probability lies between zero and one; a count may need to be a non-negative integer. Then estimate a y-range that includes the important outputs of every graphed function.

Use zoom, trace, tables and autoscale as exploration tools, but check their results. Autoscale may create an unhelpful view when one extreme value compresses the rest of the graph.

After finding solutions in one window, expand the domain or justify why no additional solutions are relevant. This is especially important for higher-degree polynomials and periodic functions.

Key terms

Graph window
The displayed x- and y-ranges on a graphing screen.
Contextual domain
Inputs permitted by the real-world meaning of the variable.
Zoom
A change of graph scale used to inspect behavior more closely.
Clipping
Failure to display part of a graph because it lies outside the selected window.

Worked examples

A window that misses both roots

Why does x2=100x^2=100 appear to have no solution in the window −5≤x≤5-5\le x\le5?

  1. 1Rewrite as x2−100=0x^2-100=0.
  2. 2The actual roots are x=±10x=\pm10.
  3. 3Both lie outside the displayed x-range.
  4. 4Expand the x-window beyond -10 and 10, then use the zero command.

Answer: The equation has x=−10x=-10 and x=10x=10; the original window hid them.

Scale a wide quadratic

Select a useful x-window for 0.001x2−2=00.001x^2-2=0.

  1. 1Estimate x2=2000x^2=2000, so ∣x∣≈44.7|x|\approx44.7.
  2. 2A standard window −10≤x≤10-10\le x\le10 is too narrow.
  3. 3A window such as −60≤x≤60-60\le x\le60 displays both roots.

Answer: The roots are x≈−44.721x\approx-44.721 and x≈44.721x\approx44.721.

Use context to remove a visible root

A time model produces graph intersections at t=−3.2t=-3.2 and t=5.7t=5.7. Which should be reported if tt measures hours after an experiment begins?

  1. 1The contextual domain is t≥0t\ge0.
  2. 2The value t=−3.2t=-3.2 lies outside that domain.
  3. 3Keep the non-negative solution and state its meaning.

Answer: Report t≈5.7t\approx5.7 hours.

Solving linear equations and systems with technology

A system of linear equations asks for values that satisfy every equation simultaneously. In two variables, the solution is an ordered pair (x,y)(x,y); in three variables, it is an ordered triple (x,y,z)(x,y,z).

AI HL requires technology for systems in up to three variables. Enter equations in a consistent variable order and standard form. A missing variable has coefficient zero, and constants belong on the correct side.

For a 2×22\times2 system, graphing the two lines makes the solution visible as an intersection. For a 3×33\times3 system, a simultaneous-equation solver is normally more efficient.

Current AI HL examination systems are designed to have a unique solution, but you must still substitute the ordered values into all original equations.

Key terms

Linear system
Two or more linear equations involving the same variables.
Simultaneous solution
Values that satisfy every equation in the system at the same time.
Coefficient
The numerical multiplier of a variable.
Unique solution
Exactly one ordered pair or triple satisfies the system.
Standard form
A consistent arrangement such as ax+by=cax+by=c or ax+by+cz=dax+by+cz=d.

Worked examples

Two-variable system

Solve 2x+3y=192x+3y=19 and x−y=2x-y=2.

  1. 1Enter the coefficient rows [2,3∣19][2,3\mid19] and [1,−1∣2][1,-1\mid2] in a simultaneous-equation solver.
  2. 2The GDC gives x=5x=5, y=3y=3.
  3. 3Check: 2(5)+3(3)=192(5)+3(3)=19 and 5−3=25-3=2.

Answer: The unique solution is (x,y)=(5,3)(x,y)=(5,3).

Three-variable system

Solve x+y+z=6x+y+z=6, 2x−y+z=32x-y+z=3, and x+2y−z=2x+2y-z=2.

  1. 1Choose a three-unknown simultaneous-equation solver.
  2. 2Enter rows [1,1,1∣6][1,1,1\mid6], [2,−1,1∣3][2,-1,1\mid3], and [1,2,−1∣2][1,2,-1\mid2].
  3. 3The output is x=1x=1, y=2y=2, z=3z=3.
  4. 4Substitute the triple into all three original equations.

Answer: The solution is (x,y,z)=(1,2,3)(x,y,z)=(1,2,3).

Include a zero coefficient

How should 2x+5z=92x+5z=9 be entered in an x,y,zx,y,z system?

  1. 1Write the equation as 2x+0y+5z=92x+0y+5z=9.
  2. 2Keep the same variable order used by every other equation.
  3. 3Enter the row [2,0,5∣9][2,0,5\mid9].

Answer: Use zero as the missing y-coefficient; do not shift the z-coefficient into the y column.

Rules and formulas

Two-variable linear system

a1x+b1y=c1,a2x+b2y=c2a_1x+b_1y=c_1,\qquad a_2x+b_2y=c_2

Three-variable linear system

aix+biy+ciz=di(i=1,2,3)a_ix+b_iy+c_iz=d_i\qquad(i=1,2,3)

Solving polynomial equations with technology

A polynomial equation can be written as anxn+an−1xn−1+⋯+a1x+a0=0a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=0, where an≠0a_n\ne0. A degree-nn polynomial has at most nn distinct real roots.

A polynomial solver normally asks for coefficients in descending powers. Enter zero for every missing power. For example, the coefficient list for 2x4−5x+7=02x^4-5x+7=0 is 2,0,0,−5,72,0,0,-5,7.

Decide whether the question needs real roots only. Some calculators display complex roots; a graph of a real function shows only real x-intercepts. AI HL problems often require interpreting real solutions in context.

Use a graph to understand the number and location of real roots, then use the polynomial or zero tool for accurate values. Check approximate roots by evaluating the polynomial.

Key terms

Polynomial
A finite sum of non-negative integer powers of x with numerical coefficients.
Degree
The highest power of x with a nonzero coefficient.
Leading coefficient
The coefficient of the highest-power term.
Real root
A real number that makes the polynomial equal zero.
Complex root
A non-real solution involving the imaginary unit; not an x-intercept on a real graph.

Worked examples

Quadratic with decimal roots

Use technology to solve 2x2+3x−6=02x^2+3x-6=0.

  1. 1Choose a degree-2 polynomial solver.
  2. 2Enter coefficients 2,3,−62,3,-6.
  3. 3The outputs are approximately −2.637458609-2.637458609 and 1.1374586091.137458609.
  4. 4Substitute both values into the original quadratic before rounding.

Answer: x≈−2.637x\approx-2.637 or x≈1.137x\approx1.137.

Cubic with three real roots

Solve x3−4x−1=0x^3-4x-1=0.

  1. 1The missing x2x^2 term has coefficient 00.
  2. 2Enter the coefficient list 1,0,−4,−11,0,-4,-1.
  3. 3The solver returns three real roots.
  4. 4Check that every result makes the polynomial approximately zero.

Answer: x≈−1.861,−0.254,2.115x\approx-1.861,-0.254,2.115.

Polynomial with no real roots

Solve x2+4=0x^2+4=0 over the real numbers.

  1. 1Enter coefficients 1,0,41,0,4.
  2. 2A real-mode solver reports no real roots; a complex-mode solver may display imaginary answers.
  3. 3Also, x2+4>0x^2+4\gt0 for every real xx.

Answer: There are no real solutions.

Rules and formulas

Polynomial standard form

anxn+an−1xn−1+⋯+a1x+a0=0,an≠0a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=0,\qquad a_n\ne0

Maximum number of distinct real roots

number of distinct real roots≤n\text{number of distinct real roots}\le n

Nonlinear equations and domain restrictions

Nonlinear equations include powers, roots, reciprocals, absolute values, exponentials and logarithms. The same graph/solver workflow applies, but restrictions become especially important.

Write restrictions before solving: denominators cannot equal zero; even roots require non-negative radicands; logarithm arguments must be positive. Transformations such as squaring or multiplying by an expression can create extraneous solutions.

A GDC may return a value outside the original domain or may fail near a discontinuity. Always substitute into the original—not merely a rearranged or squared equation.

Key terms

Nonlinear equation
An equation whose variables do not all occur only to the first power in a linear combination.
Restriction
A value excluded because an expression is undefined or the context does not permit it.
Extraneous solution
A value introduced by an algebraic transformation that does not solve the original equation.
Discontinuity
A break or undefined point in a function's graph.

Worked examples

Rational equation

Solve 1x−1=x+1\frac{1}{x-1}=x+1.

  1. 1State the restriction x≠1x\ne1.
  2. 2Graph Y1=1x−1Y_1=\frac{1}{x-1} and Y2=x+1Y_2=x+1, or solve the zero form.
  3. 3The intersections occur at x≈−1.414214x\approx-1.414214 and x≈1.414214x\approx1.414214.
  4. 4Both values satisfy the restriction and the original equation.

Answer: x=−2x=-\sqrt2 or x=2x=\sqrt2, approximately −1.414-1.414 and 1.4141.414.

Radical equation and extraneous candidate

Solve x+1=x−1\sqrt{x+1}=x-1.

  1. 1The original equation requires x≥1x\ge1 because the square root is non-negative.
  2. 2A squared/rearranged equation produces candidates x=0x=0 and x=3x=3.
  3. 3Reject x=0x=0, which violates x≥1x\ge1 and does not satisfy the original equation.
  4. 4Check x=3x=3: 4=2=3−1\sqrt4=2=3-1.

Answer: The only valid solution is x=3x=3.

Absolute-value equation

Solve ∣2x−5∣=7|2x-5|=7 using a graph.

  1. 1Graph Y1=∣2x−5∣Y_1=|2x-5| and Y2=7Y_2=7.
  2. 2Find both intersections; the V-shaped graph has two branches.
  3. 3The x-coordinates are −1-1 and 66.
  4. 4Substitute both values into the original absolute-value equation.

Answer: x=−1x=-1 or x=6x=6.

Rules and formulas

Even-root restriction

u(x) requires u(x)≥0\sqrt{u(x)}\text{ requires }u(x)\ge0

Logarithm restriction

log⁡b(u(x)) requires u(x)>0\log_b(u(x))\text{ requires }u(x)\gt0

Denominator restriction

p(x)q(x) requires q(x)≠0\frac{p(x)}{q(x)}\text{ requires }q(x)\ne0

Exponential and logarithmic equations

Exponential and logarithmic equations often have no convenient solution using elementary algebra alone. A graph, intersection command or numerical solver can find the required decimal efficiently.

For exponential models, identify the meaningful domain, often t≥0t\ge0. For logarithmic equations, enforce positive arguments before interpreting any output.

When an exact logarithmic rearrangement is available, it provides a useful check. Technology should still be used to evaluate the decimal accurately and to investigate more complicated models.

Key terms

Exponential equation
An equation in which the variable appears in an exponent.
Logarithmic equation
An equation containing a logarithm of an expression involving the variable.
Growth factor
The multiplier applied per period in an exponential growth model.
Decay factor
A multiplier between zero and one applied per period in a decay model.

Worked examples

Exponential target time

Solve 3(1.08)t=53(1.08)^t=5 for t≥0t\ge0.

  1. 1Graph Y1=3(1.08)tY_1=3(1.08)^t and Y2=5Y_2=5, or use a numerical solver.
  2. 2Choose a non-negative time window that includes the crossing.
  3. 3The GDC gives t≈6.637457t\approx6.637457.
  4. 4Check that 3(1.08)6.637457≈53(1.08)^{6.637457}\approx5.

Answer: t≈6.64t\approx6.64 periods.

Natural-logarithm equation

Solve ln⁡(x−2)=0.7\ln(x-2)=0.7.

  1. 1State the restriction x>2x\gt2.
  2. 2Graph Y1=ln⁡(x−2)Y_1=\ln(x-2) and Y2=0.7Y_2=0.7, or use a solver with an initial guess above 2.
  3. 3The numerical result is x≈4.013753x\approx4.013753.
  4. 4Check: ln⁡(4.013753−2)≈0.7\ln(4.013753-2)\approx0.7.

Answer: x≈4.014x\approx4.014.

Logarithmic equation with an invalid algebraic root

Solve log⁡10x+log⁡10(x−9)=1\log_{10}x+\log_{10}(x-9)=1.

  1. 1The domain requires x>9x\gt9.
  2. 2Combining logs leads to x(x−9)=10x(x-9)=10, whose algebraic candidates are 1010 and −1-1.
  3. 3Reject −1-1 because logarithm arguments would not be positive.
  4. 4Check x=10x=10: log⁡1010+log⁡101=1\log_{10}10+\log_{10}1=1.

Answer: The valid solution is x=10x=10.

Rules and formulas

Exponential solution check

abx=c⟹x=ln⁡(ca)ln⁡bab^x=c\Longrightarrow x=\frac{\ln(\frac{c}{a})}{\ln b}

Logarithm argument condition

log⁡bu is real only when u>0\log_b u\text{ is real only when }u\gt0

Exact versus approximate solutions

An exact solution represents the value without rounding, such as x=3x=\sqrt3 or x=74x=\frac{7}{4}. An approximate solution is a rounded decimal, such as x≈1.732x\approx1.732.

If a question asks for an exact value and the structure is recognizable, give the exact form. A numerical graph or solver cannot turn a rounded decimal back into a proven exact answer.

If the question requests a decimal or the equation has no elementary exact form, keep guard digits and round the final answer to the required significant figures or decimal places.

Context may determine the final form: money usually uses two decimal places; measurements use appropriate significant figures; counts may require a whole-number threshold rather than ordinary rounding.

Key terms

Exact value
An unrounded symbolic or numerical representation of a solution.
Approximation
A numerical value close to the exact solution.
Guard digits
Extra stored digits retained during calculations before final rounding.
Threshold
The first input value for which a stated condition becomes true.

Worked examples

Preserve an exact root

Solve x2=3x^2=3.

  1. 1A GDC gives approximately x=±1.732050808x=\pm1.732050808.
  2. 2The algebraic equation has the exact solutions x=±3x=\pm\sqrt3.
  3. 3Use the exact form unless a decimal accuracy is requested.

Answer: Exact: x=±3x=\pm\sqrt3. Approximate: x≈±1.732x\approx\pm1.732.

Round only the final solution

A solver returns x=1.324717957244…x=1.324717957244\ldots. Give the result to three significant figures.

  1. 1Keep the stored value for substitution and later calculations.
  2. 2The first three significant digits are 1, 3 and 2; the next digit is 4.
  3. 3Do not round upward.

Answer: x≈1.32x\approx1.32 to three significant figures.

Rules and formulas

Approximate equality symbol

x≈1.732x\approx1.732

Exact equality symbol

x=3x=\sqrt3

Interpreting GDC output and validating solutions

A solver message such as “no solution,” “no sign change” or “did not converge” describes the attempted computation, not necessarily the mathematics. Check the window, bounds, initial guess, mode and equation entry before drawing a conclusion.

Validate a numerical root rr by computing a residual R=∣h(r)∣R=|h(r)|, or for f(x)=g(x)f(x)=g(x), R=∣f(r)−g(r)∣R=|f(r)-g(r)|. A small residual supports the numerical accuracy but does not replace domain and context checks.

For a linear system, substitute the entire ordered pair or triple into every equation. For a polynomial, confirm all reported real roots and compare the number of roots with the graph and degree.

Interpret the sign and units. A negative time, non-integer item count, probability outside [0,1], or negative length may be mathematically computed but contextually impossible.

Key terms

Residual check
Evaluation of how close a candidate makes the original equation to equality.
Convergence failure
A numerical solver's inability to stabilize from the supplied settings.
Validation
Checking mathematics, restrictions, accuracy and context before accepting a result.
Interpretation
Explaining what a calculated value means in the original situation.

Worked examples

Residual for a cubic root

Check r=2.114907541r=2.114907541 as a root of h(x)=x3−4x−1h(x)=x^3-4x-1.

  1. 1Use the displayed candidate r=2.114907541r=2.114907541 exactly as given, rather than an unshown longer stored root.
  2. 2Substitution gives h(r)=r3−4r−1≈−4.49032×10−9h(r)=r^3-4r-1\approx-4.49032\times10^{-9}, close to zero.
  3. 3The graph also shows a crossing near 2.115.

Answer: The displayed candidate has residual approximately −4.49×10−9-4.49\times10^{-9}. It is a numerical approximation, not an exact zero; a fuller-precision stored root can have a much smaller residual.

Interpret a solver failure

A zero command reports “no sign change” for h(x)=(x−2)2h(x)=(x-2)^2 between 11 and 33. Does this prove there is no root?

  1. 1Both (h(1)) and (h(3)) equal 11, so there is no endpoint sign change.
  2. 2However, h(2)=0h(2)=0.
  3. 3The root has even multiplicity and the graph only touches the axis.

Answer: No. There is a repeated root at x=2x=2; use the graph minimum or polynomial solver.

Interpret a non-integer solution

A break-even model gives x≈26.727x\approx26.727 products. What should a business conclude?

  1. 1A fraction of a product is not permitted.
  2. 2Test the adjacent integers rather than rounding automatically.
  3. 3The model gives a loss at 26 and a profit at 27.

Answer: The minimum whole-number break-even quantity is 27 products.

Rules and formulas

Zero residual

R=∣h(r)∣R=|h(r)|

Intersection residual

R=∣f(r)−g(r)∣R=|f(r)-g(r)|

Step-by-step calculator and GDC methods

Menu names vary between TI-Nspire, TI-84, Casio ClassWiz and Casio graphing models. Learn the equivalent operations on your own approved calculator rather than memorizing one brand's key sequence.

Before any method, clear or inspect stored functions, use parentheses carefully, select real or complex mode as appropriate, and confirm angle mode when trigonometric expressions appear.

For graph zero: enter one zero function, set the window, display the graph, choose Zero/Root and bracket the crossing. For intersection: enter both functions and select the intended crossing. For tables: select a start and step that suit the search.

For polynomial and simultaneous-equation tools: select the correct degree or number of unknowns, enter coefficients in a fixed order, use zero placeholders, solve, and verify in the original equation or system.

Key terms

Equation solver
A calculator tool that searches numerically for a solution near a guess or within bounds.
Polynomial solver
A calculator tool that finds roots from an ordered coefficient list.
Simultaneous-equation solver
A tool that solves a coefficient system in two or three variables.
Trace
A graph feature that displays approximate coordinates as the cursor moves along a curve.

Worked examples

Generic GDC zero method

Find a root of x3−x−1=0x^3-x-1=0.

  1. 1Enter Y1=x3−x−1Y_1=x^3-x-1.
  2. 2Use a table or graph to identify the crossing between 11 and 22.
  3. 3Choose Zero/Root and supply left and right bounds around that crossing.
  4. 4Record x≈1.324718x\approx1.324718, then evaluate the original expression at the stored result.

Answer: x≈1.325x\approx1.325 to three decimal places.

Generic GDC intersection method

Solve 0.5x+2=8(0.7)x0.5x+2=8(0.7)^x for x≥0x\ge0.

  1. 1Enter Y1=0.5x+2Y_1=0.5x+2 and Y2=8(0.7)xY_2=8(0.7)^x.
  2. 2Set an x-window beginning at 0 and adjust the y-window until both curves are visible.
  3. 3Choose Intersect and select the visible crossing.
  4. 4Store the x-coordinate and check both function values.

Answer: x≈2.518x\approx2.518.

Generic simultaneous-system method

Solve x+y+z=6x+y+z=6, 2x−y+z=32x-y+z=3, x+2y−z=2x+2y-z=2.

  1. 1Choose a three-unknown simultaneous-equation tool.
  2. 2Enter coefficient rows [1,1,1∣6][1,1,1\mid6], [2,−1,1∣3][2,-1,1\mid3], [1,2,−1∣2][1,2,-1\mid2].
  3. 3Solve and record the ordered triple.
  4. 4Check the triple in all three equations.

Answer: (x,y,z)=(1,2,3)(x,y,z)=(1,2,3).

Common calculator mistakes and IB exam strategy

Technology questions still assess mathematical communication. Define variables, show the equation or system, state the domain or restrictions, identify the technological method, report all valid solutions, and interpret the result.

Common input mistakes include omitted brackets, missing multiplication signs, incorrect powers, coefficients in the wrong order, failure to enter zero placeholders, and graphing only one side without subtracting the other.

Common search mistakes include using a narrow window, finding only one of several roots, trusting autoscale, using an initial guess near the wrong root, and interpreting “no sign change” as proof of no solution.

Common reporting mistakes include copying the y-coordinate instead of x, rounding too early, using an equals sign for an approximation, retaining an extraneous root, ignoring units, and failing to apply an integer or non-negative constraint.

Worked examples

Diagnose a coefficient-entry error

A student enters x3−4x−1=0x^3-4x-1=0 as coefficients 1,−4,−11,-4,-1. What went wrong?

  1. 1A cubic list must represent x3,x2,x,x^3,x^2,x, and the constant in order.
  2. 2The polynomial has no x2x^2 term, so its coefficient is 00.
  3. 3The correct list is 1,0,−4,−11,0,-4,-1.

Answer: The zero placeholder for the missing square term was omitted.

Diagnose a missed root

A student reports only x=3x=3 for x2−x−6=0x^2-x-6=0 because the graph window begins at x=0x=0.

  1. 1The selected domain excludes negative x-values.
  2. 2Expand the window to include x=−2x=-2.
  3. 3Use the zero command at both crossings and check both values.

Answer: The complete solution is x=−2x=-2 or x=3x=3.

Write an exam-quality conclusion

A decay model gives t=7.951060681t=7.951060681 days for the amount to reach 3030 units. The amount is checked once at the end of each whole day.

  1. 1Do not round 7.951 to 8 without explaining the discrete interpretation.
  2. 2The threshold is crossed between days 7 and 8.
  3. 3The first scheduled whole-day check at or beyond the threshold is day 8.

Answer: The amount first reaches the required level by the end of day 8.

IB-style practice with worked solutions

Attempt each problem with your own GDC before opening the worked method. For technology questions, write enough setup that another student could reproduce the calculation on a different approved calculator.

Unless stated otherwise, give non-exact numerical answers to three significant figures and check them in the original equation.

Worked examples

Practice 1 · Linear equation

Use technology to solve 5x−13=275x-13=27.

  1. 1Enter the equation in a solver or graph Y1=5x−13Y_1=5x-13, Y2=27Y_2=27.
  2. 2The solver gives x=8x=8.
  3. 3Check: 5(8)−13=275(8)-13=27.

Answer: x=8x=8.

Practice 2 · Two-variable system

Solve x+y=14x+y=14 and 2x−y=72x-y=7.

  1. 1Enter rows [1,1∣14][1,1\mid14] and [2,−1∣7][2,-1\mid7].
  2. 2The GDC returns x=7x=7, y=7y=7.
  3. 3Substitution satisfies both equations.

Answer: (x,y)=(7,7)(x,y)=(7,7).

Practice 3 · Three-variable system

Solve x+y+z=6x+y+z=6, 2x−y+z=32x-y+z=3, x+2y−z=2x+2y-z=2.

  1. 1Choose three unknowns in the simultaneous-equation tool.
  2. 2Enter [1,1,1∣6][1,1,1\mid6], [2,−1,1∣3][2,-1,1\mid3], [1,2,−1∣2][1,2,-1\mid2].
  3. 3The output is (1,2,3)(1,2,3).
  4. 4Check the triple in all equations.

Answer: (x,y,z)=(1,2,3)(x,y,z)=(1,2,3).

Practice 4 · Quadratic roots

Solve 2x2+3x−6=02x^2+3x-6=0.

  1. 1Enter polynomial coefficients 2,3,−62,3,-6.
  2. 2The solver returns two real values.
  3. 3Substitute the stored values before rounding.

Answer: x≈−2.64x\approx-2.64 or x≈1.14x\approx1.14.

Practice 5 · Cubic roots

Find all real zeros of x3−4x−1x^3-4x-1.

  1. 1Enter coefficients 1,0,−4,−11,0,-4,-1; the zero represents the missing x2x^2 term.
  2. 2A graph confirms three real crossings.
  3. 3Check every root in the cubic.

Answer: x≈−1.86,−0.254,2.11x\approx-1.86,-0.254,2.11.

Practice 6 · Repeated root

Use technology to solve (x−2)2(x+1)=0(x-2)^2(x+1)=0 and describe the graph at each root.

  1. 1Graph the expanded or factored polynomial.
  2. 2It crosses the axis at x=−1x=-1.
  3. 3It touches and turns at x=2x=2, indicating even multiplicity.

Answer: Roots: −1-1 and 22; 22 is a double root.

Practice 7 · Exponential equation

Solve 4(1.12)t=104(1.12)^t=10 for t≥0t\ge0.

  1. 1Graph the exponential and the horizontal line y=10y=10, or use a numerical solver.
  2. 2The intersection gives t≈8.085249815t\approx8.085249815.
  3. 3Check the stored value in the original equation.

Answer: t≈8.09t\approx8.09 periods.

Practice 8 · Logarithmic restriction

Solve log⁡10x+log⁡10(x−9)=1\log_{10}x+\log_{10}(x-9)=1.

  1. 1State x>9x\gt9.
  2. 2Technology or algebra gives candidates x=10x=10 and x=−1x=-1.
  3. 3Reject -1 because the logarithms are undefined.

Answer: x=10x=10.

Practice 9 · Radical validation

Solve x+1=x−1\sqrt{x+1}=x-1.

  1. 1The original equation requires x≥1x\ge1.
  2. 2A solver on the original graphs, or a squared equation, suggests x=3x=3; squaring may also suggest x=0x=0.
  3. 3Substitution rejects 0 and confirms 3.

Answer: x=3x=3.

Practice 10 · Table and bracket

Locate and solve x3−x−1=0x^3-x-1=0.

  1. 1A table gives h(1.3)≈−0.103h(1.3)\approx-0.103 and h(1.4)≈0.344h(1.4)\approx0.344.
  2. 2Use 1.3 and 1.4 as a bracket for a zero command.
  3. 3Check the result in the original cubic.

Answer: x≈1.32x\approx1.32 to three significant figures.

Practice 11 · Repair the graph window

A graph of x2−100x^2-100 on −5≤x≤5-5\le x\le5 shows no roots. Explain and solve.

  1. 1The displayed x-range excludes both intercepts.
  2. 2Expand the window beyond -10 and 10.
  3. 3Use the zero command at each crossing.

Answer: x=−10x=-10 or x=10x=10.

Practice 12 · Contextual break-even

For R(x)=1200xR(x)=1200x and C(x)=30000+15x1.5C(x)=30000+15x^{1.5}, x≥0x\ge0, find the first whole number of units that is profitable.

  1. 1Graph R and C and find their intersection.
  2. 2The continuous intersection is x≈26.727x\approx26.727.
  3. 3Evaluate profit at the adjacent whole numbers.
  4. 4At 26 units profit is negative; at 27 it is positive.

Answer: The first profitable whole-number output is 27 units.

Practice 13 · Interpret solver output

A solver returns x=1.414213562x=1.414213562 and x=−1.414213562x=-1.414213562 for 1x−1=x+1\frac1{x-1}=x+1. Validate and report the answers.

  1. 1The restriction is x≠1x\ne1, so both candidates remain possible.
  2. 2Multiply the original equation by x−1≠0x-1\ne0: 1=(x−1)(x+1)=x2−11=(x-1)(x+1)=x^2-1, so x2=2x^2=2.
  3. 3Therefore x=±2x=\pm\sqrt2. Both satisfy x≠1x\ne1 and substitute into the original equation; their decimals match the solver output.

Answer: x=±2x=\pm\sqrt2, approximately ±1.414\pm1.414.

Practice 14 · Full IB-style decay decision

A substance has amount A(t)=120(0.84)tA(t)=120(0.84)^t milligrams after tt days. It is safe when the amount is at most 3030 mg. Measurements occur at the end of each whole day. Determine the first safe measurement day.

  1. 1Solve 120(0.84)t=30120(0.84)^t=30 using an intersection or numerical solver with t≥0t\ge0.
  2. 2The continuous threshold occurs at t≈7.951060681t\approx7.951060681.
  3. 3Because measurements occur only at whole days, compare days 7 and 8 rather than rounding without interpretation.
  4. 4A(7)>30A(7)\gt30 while A(8)≤30A(8)\le30.
  5. 5State the result with units and connect it to the measurement schedule.

Answer: The first safe scheduled measurement is at the end of day 8.

Topic summary and cheat sheet

Define the mathematical equation before opening a calculator. Decide whether a zero, intersection, table, polynomial solver or simultaneous-equation solver is the clearest representation.

Convert f(x)=g(x)f(x)=g(x) into h(x)=f(x)−g(x)=0h(x)=f(x)-g(x)=0 whenever a zero method is convenient.

For graphs, choose a meaningful domain and window, locate every crossing or tangency, then use a numerical command for accuracy.

For systems and polynomials, preserve variable/power order and enter zero coefficients for missing terms.

Treat outputs as candidates: check the original equation, restrictions, requested accuracy and context before reporting them.

Rules and formulas

Zero and intersection equivalence

f(x)=g(x)  ⟺  f(x)−g(x)=0f(x)=g(x)\iff f(x)-g(x)=0

Root definition

h(r)=0h(r)=0

Sign-change bracket

h(a)h(b)<0h(a)h(b)\lt0

Residual

R=∣LHS−RHS∣R=|\text{LHS}-\text{RHS}|

Polynomial form

anxn+⋯+a1x+a0=0a_nx^n+\cdots+a_1x+a_0=0

Two-variable system

a1x+b1y=c1,a2x+b2y=c2a_1x+b_1y=c_1,\quad a_2x+b_2y=c_2

Topic summary

  • A solution makes the original equation or every equation in a system true.
  • AI HL explicitly uses technology for linear systems in up to three variables and polynomial equations.
  • A root or zero rr satisfies h(r)=0h(r)=0 and appears as the x-intercept (r,0)(r,0).
  • Solving f(x)=g(x)f(x)=g(x) by intersections is equivalent to solving f(x)−g(x)=0f(x)-g(x)=0.
  • A numerical solver may require an initial guess or bracket and may find only one of several roots.
  • Use graphs to understand the number and location of solutions, then use a numerical command for accuracy.
  • Use tables and sign changes to bracket crossing roots, while remembering that touching roots may not change sign.
  • A graph window displays only part of a graph; it is not proof about the full domain.
  • Choose the x-domain from algebra and context, then choose a y-window that shows all relevant functions.
  • In linear systems, keep the same variable order in every row and enter zero for a missing variable.
  • In polynomial solvers, enter coefficients in descending powers and include zero placeholders for missing powers.
  • A degree-n polynomial has at most n distinct real roots, but some roots may be repeated or non-real.
  • Rational, radical and logarithmic equations require domain restrictions before solutions are accepted.
  • Squaring or clearing denominators can create extraneous candidates, so check the original equation.
  • Use exact forms when required and available; otherwise report approximate values with the requested accuracy.
  • Keep guard digits for checking and subsequent calculations, and round only the final reported answer.
  • A small residual supports numerical accuracy but does not replace domain or contextual validation.
  • A solver failure may indicate poor bounds, window, mode or input rather than absence of a solution.
  • For counts and scheduled times, test adjacent integers and interpret thresholds instead of rounding mechanically.
  • An IB-quality response shows the model, technological method, all valid solutions, checks and contextual conclusion.