Graphical/numerical solutions | Year 9

01 / LEARN

Understand the idea.

A root is where the output is zero

A graphical solution is an input where two plotted outputs agree. For $x^{2} = 5$ , intersect $y = x^{2}$ with $y = 5$ , or find the zeros of $y = x^{2} - 5$ . The solutions are the intersections' horizontal coordinates.

A graph gives estimates whose accuracy depends on its scale. Use a table or calculator to refine them, and write $\approx$ for rounded values. Inspect the full stated domain so a second intersection is not missed.

Bracket and improve an estimate

To bracket a root, find inputs on either side with outputs of opposite signs. A continuous polynomial cannot move from a negative output to a positive one without passing through zero somewhere between them.

Test inside the interval and retain a smaller sign-changing interval. Continue until every value within the bracket rounds to the same required answer. Check other intervals too: a sign change locates a root, but does not automatically list all roots.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example

Estimate both solutions of $x^{2} = 5$ to two decimal places.

Worked solution

1
Locate the positive root roughly.
$2^{2} < 5 < 3^{2}$
The positive solution lies between 2 and 3; the upward parabola also has a negative solution.
2
Test close decimal bounds.
$\begin{aligned} {2.235}^{2} & = 4.995225 < 5 \\ {2.237}^{2} & = 5.004169 > 5 \end{aligned}$
Continuity and increasing squares for positive inputs bracket the positive root.
3
Round the bracketed root.
$2.235 < x < 2.237 \Rightarrow x \approx 2.24$
Every number in this interval rounds to 2.24 at two decimal places.
4
Include the symmetric solution.
$x \approx - 2.24 or x \approx 2.24$
Squaring a number and its negative gives the same result, so both branches are required.

Answer

$x \approx - 2.24$ or $2.24$

EXAMPLE 02

Worked example

Use substitution to locate the positive root of $x^{2} - 2 x - 1 = 0$ between two consecutive tenths.

LaTeX graph of y equals x squared minus two x minus one, with x from minus four to six and y from minus six to twelve. Horizontal grid spacing is one unit and vertical grid spacing is two units. No root estimate is marked.

Worked solution

1
Evaluate at the lower tenth.
$f (2.4) = {2.4}^{2} - 2 (2.4) - 1 = 5.76 - 4.8 - 1 = - 0.04$
This input gives a negative output.
2
Evaluate at the next tenth.
$f (2.5) = {2.5}^{2} - 2 (2.5) - 1 = 6.25 - 5 - 1 = 0.25$
This input gives a positive output.
3
State the root interval.
$2.4 < x < 2.5$
The polynomial is continuous and changes sign between these consecutive tenths.
4
Confirm which root is being located.
$x^{2} - 2 x - 1 = {(x - 1)}^{2} - 2$
This parabola increases to the right of its axis x equal to 1, so the bracket identifies its unique positive root.

Answer

Between $2.4$ and $2.5$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

For $f (x) = x^{2} - 3$ , $f (1.7) = - 0.11$ and $f (1.8) = 0.24$ . Where is a positive root?

Show answer & reasoning

Between $1.7$ and $1.8$ .

The polynomial is continuous, and $f (1.7) < 0 < f (1.8)$ . It must cross zero for some input strictly between $1.7$ and $1.8$ ; these non-zero endpoint values exclude the endpoints.

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Use a calculator to estimate both roots of $x^{2} = 7$ to one decimal place.
2 marks
For $f (x) = x^{2} - 6$ , calculate $f (2.4)$ and $f (2.5)$ .
2 marks
Explain what the previous outputs show about a root of $x^{2} - 6 = 0$ .
2 marks
The graph of $y = {(x - 1)}^{2} - 2$ is supplied. Estimate its roots to the nearest half unit.
2 marks
Use a table to locate the positive root of $x^{2} - 8 = 0$ between consecutive tenths.
3 marks
Why does finding $x \approx 2.45$ not fully solve $x^{2} = 6$ over the real numbers?
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Use a calculator to estimate both roots of $x^{2} = 7$ to one decimal place.
2 marks
$\sqrt{7} \approx 2.64575$ , which rounds to $2.6$ .
$x \approx - 2.6, 2.6$
For $f (x) = x^{2} - 6$ , calculate $f (2.4)$ and $f (2.5)$ .
2 marks
${2.4}^{2} - 6 = 5.76 - 6 = - 0.24$ .
${2.5}^{2} - 6 = 6.25 - 6 = 0.25$ .
$- 0.24$ and $0.25$
Explain what the previous outputs show about a root of $x^{2} - 6 = 0$ .
2 marks
The continuous quadratic changes from negative to positive.
A positive root lies between $2.4$ and $2.5$ .
The graph of $y = {(x - 1)}^{2} - 2$ is supplied. Estimate its roots to the nearest half unit.
2 marks
The actual crossings are near $- 0.414$ and $2.414$ .
The nearest half-unit readings are $- 0.5$ and $2.5$ .
About $- 0.5$ and $2.5$ .
Use a table to locate the positive root of $x^{2} - 8 = 0$ between consecutive tenths.
3 marks
${2.8}^{2} - 8 = - 0.16$ and ${2.9}^{2} - 8 = 0.41$ .
Between $2.8$ and $2.9$ .
Why does finding $x \approx 2.45$ not fully solve $x^{2} = 6$ over the real numbers?
3 marks
Both a positive and its negative have the same square.
Check both sides of the parabola.
There is also $x \approx - 2.45$ .

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Curriculum reference & lesson scope

Australian Curriculum v9 / SA Curriculum Mathematics

AC9M9A02, AC9M9A04–A06: monic quadratic expressions, integer-root equations, graphs and models.

Official curriculum source ↗

In this lesson

Understand the idea.

A root is where the output is zero

Bracket and improve an estimate

See the method in action.

Worked example

Locate the positive root roughly.

Test close decimal bounds.

Round the bracketed root.

Include the symmetric solution.

Worked example

Evaluate at the lower tenth.

Evaluate at the next tenth.

State the root interval.

Confirm which root is being located.

Try it yourself.

Check your understanding

Keep practising