Numerical Methods with CAS calculator
Use numerical methods for Victorian Year 11 Mathematical Methods (VCAA) — finding approximate solutions of equations with technology when an exact method is not practical.
You will learn to locate a solution between two values using a sign change, use a CAS calculator to solve equations to a required accuracy, and read how many solutions an equation has from its graph.
Theory
In Year 11 Mathematical Methods (Unit 2), many equations cannot be solved exactly by algebra. A numerical method finds an approximate solution: rearrange the equation to the root-finding form \(f(x)=0\), use a sign change to trap a root in an interval, then read the value from a CAS calculator to the required number of decimal places. This page shows how to bracket, solve and round.
A numerical solution of an equation is an approximate value of \(x\), found by calculation rather than exact algebra. It is stated correct to a set number of decimal places.
To use these methods, first write the equation in root-finding form \(f(x)=0\). A value of \(x\) that makes \(f(x)=0\) is a root (or solution); on the graph of \(y=f(x)\) it is an \(x\)-intercept.
The sign-change test locates a root. If \(f\) is continuous and \(f(a)\) and \(f(b)\) have opposite signs, then the curve must cross the \(x\)-axis somewhere between \(a\) and \(b\), so a root lies in the interval \((a,\,b)\). This is the intermediate value idea.
A CAS calculator then finds the root: its solve command, or graphing the function and using the zero / intersection tool, returns the value to many digits, which you round.
Root-finding form. Move everything to one side so the equation reads
Sign-change (bracketing) test. For a continuous \(f\):
Two-graph form. An equation \(g(x)=h(x)\) can instead be read as an intersection: each solution is an \(x\)-value where the graphs \(y=g(x)\) and \(y=h(x)\) meet.
Solving an equation numerically with CAS
- Rearrange: write the equation as \(f(x)=0\) by moving every term to one side (or keep it as \(g(x)=h(x)\) for the intersection method).
- Bracket: evaluate \(f\) at convenient values (often consecutive integers) and find where the sign changes — that interval contains a root.
- Solve: use the CAS solve command, or graph \(y=f(x)\) and use the zero tool (or intersect the two graphs), taking the root in your bracketing interval.
- Round: state the answer correct to the required number of decimal places, with units where the problem has them.
Write in root-finding form:
| \(f(x)\) | \(=\) | \(x^3+x-3\) |
Evaluate at consecutive integers:
| \(f(1)\) | \(=\) | \((1)^3+(1)-3\) |
| \(=\) | \(-1\) | |
| \(f(2)\) | \(=\) | \((2)^3+(2)-3\) |
| \(=\) | \(8+2-3\) | |
| \(=\) | \(7\) |
Now compare the signs of the two values.
Apply the sign-change test:
| \(f(1)\) | \(<\) | \(0\) |
| \(f(2)\) | \(>\) | \(0\) |
\(f\) is continuous and changes sign, so it must cross zero between \(x=1\) and \(x=2\).
The solution lies in the interval \((1,\,2)\).
Write in root-finding form:
| \(f(x)\) | \(=\) | \(x^3-5x+1\) |
Bracket the largest root:
| \(f(2)\) | \(=\) | \((2)^3-5(2)+1\) |
| \(=\) | \(8-10+1\) | |
| \(=\) | \(-1\) | |
| \(f(3)\) | \(=\) | \((3)^3-5(3)+1\) |
| \(=\) | \(27-15+1\) | |
| \(=\) | \(13\) |
\(f(2)<0\) and \(f(3)>0\), so the largest root lies in \((2,\,3)\).
Solve numerically with CAS:
| \(x\) | \(=\) | \(2.12842\ldots\) |
Round the \(3\)rd decimal place: the \(4\)th digit is \(4<5\), so it stays.
The largest solution is \(x=2.128\).
Rearrange to root-finding form:
| \(2^x+x\) | \(=\) | \(8\) |
| \(2^x+x-8\) | \(=\) | \(0\) |
| \(f(x)\) | \(=\) | \(2^x+x-8\) |
Bracket the root:
| \(f(2)\) | \(=\) | \(2^2+2-8\) |
| \(=\) | \(4+2-8\) | |
| \(=\) | \(-2\) | |
| \(f(3)\) | \(=\) | \(2^3+3-8\) |
| \(=\) | \(8+3-8\) | |
| \(=\) | \(3\) |
The sign changes on \((2,\,3)\), and \(f\) is increasing, so this is the only solution.
Solve numerically with CAS:
| \(x\) | \(=\) | \(2.46784\ldots\) |
Round to \(3\) decimal places: the \(4\)th digit is \(8\ge5\), so round the \(7\) up.
The solution is \(x=2.468\).
Set up the equation:
| \(x(20-2x)(30-2x)\) | \(=\) | \(800\) |
Bracket the smallest root \((0
| \(V(1)\) | \(=\) | \(1(18)(28)\) |
| \(=\) | \(504\) | |
| \(V(2)\) | \(=\) | \(2(16)(26)\) |
| \(=\) | \(832\) |
\(V(1)<800 Solve numerically with CAS: Round to \(2\) decimal places: the \(3\)rd digit is \(5\ge5\), so round the \(7\) up. The smallest value is \(x=1.88\text{ cm}\).
\(x\) \(=\) \(1.87573\ldots\)
Common pitfalls
Frequently asked questions
What is a numerical method for solving an equation?
A way to find an approximate solution when exact algebra will not work. You rearrange to \(f(x)=0\), find where the curve crosses the \(x\)-axis, and read the value from a CAS calculator to a set number of decimal places.
How do you know a solution lies between two x-values?
Use the sign-change test: if \(f\) is continuous and \(f(a)\) and \(f(b)\) have opposite signs, the curve must cross zero between them, so a root lies in \((a,\,b)\).
How do you solve an equation with a CAS calculator?
Rearrange to \(f(x)=0\), then use the solve command, or graph \(y=f(x)\) and use the zero tool. For \(g(x)=h(x)\) you can instead graph both sides and use intersection.
How many solutions does an equation have?
Graph the function \(f(x)\) (or both sides of the equation) and count the \(x\)-intercepts, or the intersection points, of the graphs. Each one is a solution.
How do you round to a given number of decimal places?
Look at the digit just after the last one you keep. If it is \(5\) or more, round up; if it is less than \(5\), leave the last digit unchanged.
Why rearrange an equation so it equals zero?
So solving becomes finding the \(x\)-intercepts (roots) of one function. The sign-change test and the CAS zero tool both work on \(f(x)=0\).