Teach Me Maths

Linear and Quadratic Equations in Context

How to form and solve linear and quadratic equations from contexts, choose a method and interpret valid solutions.

Define the variable and form an equation

State what the variable represents, including its units. Translate the relationship into an equation before solving it. The answer must then satisfy both the equation and the original context.

For a quadratic, rearrange to:

ax^2+bx+c=0

Factorise when factors are convenient. Otherwise use completing the square or the quadratic formula:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Keep both algebraic roots until you check the context.

Worked example 1

A taxi fare is £3.80 plus £1.60 per kilometre. The total is £18.20. Find the distance travelled.

  1. Let the distance be d km. Then 3.80+1.60d=18.20

  2. Your turn. Subtract the fixed charge. How many pounds pay for the distance?

    1.60d=14.40

  3. Divide by 1.60: d=9. Check 3.80+1.60(9)=18.20
  4. Answer: 9 km

Worked example 2

A rectangle has width x cm and length (x+7) cm. Its area is 60 cm². Find its dimensions.

  1. Area gives x(x+7)=60, so x^2+7x-60=0
  2. Factorise: (x+12)(x-5)=0. Thus x=-12 or x=5

  3. Your turn. Which root can represent the width in centimetres?

    Use x=5 because a width must be positive.

  4. Answer: width 5 cm, length 12 cm. Check 5\times12=60

Worked example 3

Two consecutive positive integers have product 132. Find the integers.

  1. Let the smaller integer be n. The next is n+1, so n(n+1)=132

  2. Your turn. Factorising gives (n+12)(n-11)=0. What is the positive value of n?

    n=11

  3. Answer: 11 and 12. The other algebraic value, n=-12, would give two negative integers, which the question excludes.

Use the relationship that matches the measurements

Lengths, areas and volumes produce different equations. For a right-angled triangle, identify the hypotenuse before applying Pythagoras. Do not assume that a sketch is drawn to scale.

Worked example 4

A right-angled triangle has shorter sides x cm and (x+7) cm, and hypotenuse 17 cm. Find the shorter sides.

  1. Pythagoras gives x^2+(x+7)^2=17^2. Expanding and simplifying gives x^2+7x-120=0

  2. Your turn. From (x+15)(x-8)=0, choose the positive value of x

    x=8

  3. Answer: 8 cm and 15 cm. Check 8^2+15^2=289=17^2

Worked example 5

A rectangle has width x cm and length (2x+3) cm. Its area is 40 cm². Find its dimensions to 2 decimal places.

  1. Form x(2x+3)=40, so 2x^2+3x-40=0. Use a=2, b=3, c=-40

  2. Your turn. Calculate b^2-4ac=3^2-4(2)(-40)

    The discriminant is 329

  3. x=\frac{-3\pm\sqrt{329}}4. Keep the positive root x\approx3.784589. Use this unrounded value in 2x+3
  4. Answer: width 3.78 cm and length 10.57 cm, to 2 decimal places.

Worked example 6

A model gives the height of a ball as h=-t^2+8t+3 metres, t seconds after release. Find when it first reaches the ground, giving an exact answer.

  1. At the ground h=0. Completing the square gives h=19-(t-4)^2

  2. Your turn. At ground level, what does (t-4)^2 equal?

    (t-4)^2=19

  3. Thus t=4\pm\sqrt{19}. Since \sqrt{19}>4, the minus choice is before release and must be rejected.
  4. Answer: t=4+\sqrt{19} seconds, approximately 8.36 seconds after release.

Common mistakes

  • Solving before modelling the situation. Define the variable and form an equation from the given relationships.
  • Using the zero-product rule too soon. Make one side zero before factorising and setting each factor to zero.
  • Dropping a root without explanation. Find all roots, then explain any rejection using the context.
  • Keeping impossible values. Check lengths are positive and times lie in the stated interval.
  • Rounding intermediate lengths. Keep exact values or full calculator precision until the final answer.

Now try it: Linear and Quadratic Equations in Context practice questions

More on this topic: Linear and Quadratic Equations in Context worksheet with full solutions