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Linear and quadratic simultaneous equations — 0606

Substitute the line into the curve and retain both intersections.

Notes and key methods

For a simultaneous linear and quadratic system, solve the linear equation for a variable and substitute into the other equation. The resulting quadratic may have two, one or no real roots.

Every retained x-value needs its matching y-value. When finding an area between curves, integrate upper minus lower between intersection coordinates; a negative signed integral is not a negative geometric area.

Check the conditions

Three worked examples

Example 1 · A line and a quadratic curve

A curve is \(y = x^{2} - 7 x + 25\) and a line is \(y = 3 x + 4\).

  1. Find both intersection points.
  2. Explain which graph is upper between the intersections.
  3. Find the exact enclosed area.
  4. Find the area between the two graphs.
Show the reasoning

(a)

\(\left(x - 7\right) \left(x - 3\right) = \left(x - 7\right) \left(x - 3\right)\); solve and substitute each root into the line.

Answer: \(\left\{\left( 3, \ 13\right), \left( 7, \ 25\right)\right\}\)

(b)

Between the two roots, (x−u)(x−v)<0, so the curve lies below the line.

Answer: \(3 x + \left(x - 7\right) \left(x - 3\right) + 4 < 3 x + 4\) on \(\left(3, 7\right)\)

(c)

\(\int\limits_{3}^{7} \left(- \left(x - 7\right) \left(x - 3\right)\right)\, dx\)=\(\displaystyle \frac{32}{3}\)

Answer: \(\displaystyle \frac{32}{3}\)

(d)

Integrate upper minus lower between the intersections.

\(\int\limits_{3}^{7} \left(- \left(x - 7\right) \left(x - 3\right)\right)\, dx\)=\(\displaystyle \frac{32}{3}\)

Answer: \(\displaystyle \frac{32}{3}\).

Example 2 · A line and a quadratic curve

A curve is \(y = x^{2} - 12 x + 41\) and a line is \(y = x + 5\).

  1. Find both intersection points.
  2. Explain which graph is upper between the intersections.
  3. Find the exact enclosed area.
  4. Find the area between the two graphs.
Show the reasoning

(a)

\(\left(x - 9\right) \left(x - 4\right) = \left(x - 9\right) \left(x - 4\right)\); solve and substitute each root into the line.

Answer: \(\left\{\left( 4, \ 9\right), \left( 9, \ 14\right)\right\}\)

(b)

Between the two roots, (x−u)(x−v)<0, so the curve lies below the line.

Answer: \(x + \left(x - 9\right) \left(x - 4\right) + 5 < x + 5\) on \(\left(4, 9\right)\)

(c)

\(\int\limits_{4}^{9} \left(- \left(x - 9\right) \left(x - 4\right)\right)\, dx\)=\(\displaystyle \frac{125}{6}\)

Answer: \(\displaystyle \frac{125}{6}\)

(d)

Integrate upper minus lower between the intersections.

\(\int\limits_{4}^{9} \left(- \left(x - 9\right) \left(x - 4\right)\right)\, dx\)=\(\displaystyle \frac{125}{6}\)

Answer: \(\displaystyle \frac{125}{6}\).

Example 3 · A line and a quadratic curve

A curve is \(y = x^{2} - 10 x + 41\) and a line is \(y = 2 x + 6\).

  1. Find both intersection points.
  2. Explain which graph is upper between the intersections.
  3. Find the exact enclosed area.
  4. Find the area between the two graphs.
Show the reasoning

(a)

\(\left(x - 7\right) \left(x - 5\right) = \left(x - 7\right) \left(x - 5\right)\); solve and substitute each root into the line.

Answer: \(\left\{\left( 5, \ 16\right), \left( 7, \ 20\right)\right\}\)

(b)

Between the two roots, (x−u)(x−v)<0, so the curve lies below the line.

Answer: \(2 x + \left(x - 7\right) \left(x - 5\right) + 6 < 2 x + 6\) on \(\left(5, 7\right)\)

(c)

\(\int\limits_{5}^{7} \left(- \left(x - 7\right) \left(x - 5\right)\right)\, dx\)=\(\displaystyle \frac{4}{3}\)

Answer: \(\displaystyle \frac{4}{3}\)

(d)

Integrate upper minus lower between the intersections.

\(\int\limits_{5}^{7} \left(- \left(x - 7\right) \left(x - 5\right)\right)\, dx\)=\(\displaystyle \frac{4}{3}\)

Answer: \(\displaystyle \frac{4}{3}\).

Common mistakes

Practise with answer checking

Original exam-style problems

Write complete solutions before opening the worked solution. The marking guidance rewards reasoning as well as answers.

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