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Year 9 · Quadratics and Parabolas · 16 min

Graphical/numerical solutions

Use graphs and improving numerical estimates when roots are not simple integers.

01 / LEARN

The idea, step by step

A root is where the output is zero

To solve x2=5 graphically, plot y=x2 and the horizontal line y=5. The x coordinates of their intersections are the solutions. Equivalently, plot y=x2−5 and find its x-intercepts.

A sketch gives estimates, so quote only the accuracy justified by its scale. A graph with one-unit grid squares does not justify six decimal places. A calculator or table can refine a graphical estimate.

Bracket and improve an estimate

For f⁡(x)=x2−5, f⁡(2.2)=−0.16 while f⁡(2.3)=0.29. Because this polynomial graph is continuous, a root lies between 2.2 and 2.3. Test a value inside the interval and keep the smaller interval containing a sign change.

Repeat until the required rounding is clear. Also check the other part of the graph: x2=5 has a negative solution as well as a positive one. A method must find all solutions in the stated domain.

A common trap

A graph estimate is automatically an exact answer.

Instead: A plotted intercept is approximate unless an exact calculation establishes it. State sensible rounding.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

Estimate both solutions of x2=5 to two decimal places.

  1. The graph suggests roots near −2.2 and 2.2.
  2. 2.2352=4.995225 and 2.2372=5.004169, so the positive root lies between them.
  3. Both bounds round to 2.24 to two decimal places; symmetry gives −2.24.
x≈−2.24 or 2.24
Example 2

Worked example

Use substitution to locate the positive root of x2−2⁢x−1=0 between two consecutive tenths.

  1. At 2.4, 5.76−4.8−1=−0.04.
  2. At 2.5, 6.25−5−1=0.25.
  3. The continuous graph crosses zero between those inputs.
Between 2.4 and 2.5.

Check your understanding

For f⁡(x)=x2−3, f⁡(1.7)=−0.11 and f⁡(1.8)=0.24. Where is a positive root?

Show answer

Between 1.7 and 1.8.

The continuous graph changes sign between those inputs.

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Read the questions online
  1. Use a calculator to estimate both roots of x2=7 to one decimal place.
    2 marks
  2. For f⁡(x)=x2−6, calculate f⁡(2.4) and f⁡(2.5).
    2 marks
  3. Explain what the previous outputs show about a root of x2−6=0.
    2 marks
  4. The graph of y=(x−1)2−2 is supplied. Estimate its roots to the nearest half unit.
    2 marks
  5. Use a table to locate the positive root of x2−8=0 between consecutive tenths.
    3 marks
  6. Why does finding x≈2.45 not fully solve x2=6 over the real numbers?
    3 marks
Read the worked solutions

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

  1. Use a calculator to estimate both roots of x2=7 to one decimal place.
    2 marks

    7≈2.64575, which rounds to 2.6.

    x≈−2.6,2.6
  2. For f⁡(x)=x2−6, calculate f⁡(2.4) and f⁡(2.5).
    2 marks

    2.42−6=5.76−6=−0.24.

    2.52−6=6.25−6=0.25.

    −0.24 and 0.25
  3. Explain what the previous outputs show about a root of x2−6=0.
    2 marks

    The continuous quadratic changes from negative to positive.

    A positive root lies between 2.4 and 2.5.
  4. 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.
  5. Use a table to locate the positive root of x2−8=0 between consecutive tenths.
    3 marks

    2.82−8=−0.16 and 2.92−8=0.41.

    Between 2.8 and 2.9.
  6. Why does finding x≈2.45 not fully solve x2=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≈−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 ↗
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