Applications of Quadratics
Every question with a fully worked solution.
- Applications of Quadratics - Video - Solving Quadratic Equations by Factoring Watch
Theory
To solve a problem with a quadratic, name the unknown, write the other amounts in terms of it, form an equation, solve it, then check which answers make sense. This Year 10 maths page covers number and formula problems, lengths and areas, Pythagoras, height models and the greatest height, and price and income models.
A quadratic model is an equation in which the unknown is squared, usually because two amounts that both depend on the unknown are multiplied: length times width, price times quantity, or one age times another.
Choose one unknown and write every other amount in terms of it. If two numbers add to \(23\), call them \(x\) and \(23-x\). If one length is \(14\) more than another, call them \(x\) and \(x+14\).
The quadratic usually has two solutions. Each must be tested against the situation. Reject a negative length, time, speed or count, a fraction of a person, or a value that makes another length negative.
Sometimes both solutions are valid: two different prices can give the same income, and a ball passes a given height on the way up and on the way down. If the equation has no real solutions (the discriminant is negative), the situation never happens.
Relationships that often produce quadratics:
| Situation | Relationship | In terms of \(x\) |
|---|---|---|
| Consecutive integers | \(n,\ n+1\) | even or odd: \(n,\ n+2\) or \(2n-1,\ 2n+1\) |
| Area of a rectangle | \(A=lw\) | perimeter \(P\): sides \(x\) and \(\dfrac{P}{2}-x\) |
| Triangle, trapezium, kite | \(\dfrac{1}{2}bh\), \(\dfrac{1}{2}(a+b)h\), \(\dfrac{1}{2}xy\) | multiply both sides by \(2\) first |
| Right-angled triangle | \(a^2+b^2=c^2\) | sides such as \(x\), \(x+2\) and \(x+4\) |
| Height of a ball | \(h=ut-5t^2\) | lands when \(h=0\); the top is halfway between the two landing times |
| Income | price \(\times\) number sold | \((p+ak)(N-bk)\) |
How to solve a quadratic word problem
- Define the unknown in words, with units: "let \(x\) be the width in metres".
- Express every other amount in terms of \(x\). A labelled diagram helps.
- Form the equation from the fact the question gives (an area, a total, a height).
- Rearrange to \(ax^2+bx+c=0\), dividing by any common factor, then solve by factorising, the formula or completing the square.
- Reject solutions that do not fit, and say why.
- Answer the question asked, with units. It may ask for the other side, the price, or the hypotenuse, not \(x\).
Other question types
- Height models: for \(h=1+12t-5t^2\), set \(h\) equal to the height asked for. To show a height is never reached, show \(b^2-4ac\lt 0\): for \(h=9\), \(5t^2-12t+8=0\) has \(\Delta=144-160=-16\).
- Fencing with a wall: with \(L\) m of fence for three sides, the sides are \(x\), \(x\) and \(L-2x\), so the area is \(x(L-2x)\).
- Ages: add the same number of years to every person's age before multiplying.
- A number and its reciprocal: \(x+\dfrac{1}{x}=s\); multiply every term by \(x\).
(i) Two consecutive positive even numbers have a product of \(168\). Find them.
(ii) When \(n\) people each shake hands once with everyone else, there are \(\dfrac{n(n-1)}{2}\) handshakes. How many people make \(66\) handshakes?
(i) Let the smaller number be \(n\). The next even number is \(n+2\).
| \(n(n+2)\) | \(=\) | \(168\) |
| \(n^2+2n-168\) | \(=\) | \(0\) |
| \((n+14)(n-12)\) | \(=\) | \(0\) |
| \(n\) | \(=\) | \(12 \text{ or } n=-14\) |
The numbers are positive, so reject \(-14\). The numbers are \(\boldsymbol{12}\) and \(\boldsymbol{14}\).
(ii) Substitute \(66\) for the number of handshakes and multiply both sides by \(2\).
| \(n^2-n\) | \(=\) | \(132\) |
| \(n^2-n-132\) | \(=\) | \(0\) |
| \((n-12)(n+11)\) | \(=\) | \(0\) |
A number of people cannot be negative, so there are \(\boldsymbol{12}\) people. Check: \(\dfrac{12\times 11}{2}=66\).
(i) A right-angled triangle has a hypotenuse of \(26\) cm, and one shorter side is \(14\) cm longer than the other. Find the two shorter sides.
(ii) A rectangle has a perimeter of \(46\) m and an area of \(120\ \text{m}^2\). Find its dimensions.
(i) Let the shortest side be \(x\) cm, as in the diagram.
| \(x^2+(x+14)^2\) | \(=\) | \(26^2\) |
| \(2x^2+28x+196\) | \(=\) | \(676\) |
| \(x^2+14x-240\) | \(=\) | \(0\) |
| \((x+24)(x-10)\) | \(=\) | \(0\) |
A length is positive, so \(x=10\). The sides are \(\boldsymbol{10}\) cm and \(\boldsymbol{24}\) cm. Check: \(10^2+24^2=676=26^2\).
(ii) Length \(+\) width is half the perimeter, \(23\) m. Let the width be \(x\) m, so the length is \((23-x)\) m.
| \(x(23-x)\) | \(=\) | \(120\) |
| \(x^2-23x+120\) | \(=\) | \(0\) |
| \((x-8)(x-15)\) | \(=\) | \(0\) |
Both values give the same rectangle: \(\boldsymbol{15}\) m by \(\boldsymbol{8}\) m.
(i) When does it land?
(ii) Find its greatest height.
(iii) When is it \(15\) m above the ground?
(iv) A second ball follows \(h=kt-5t^2\), where \(k\gt 0\), and its greatest height is \(5\) m. Find \(k\).
(i) Set \(h=0\) and take out the common factor.
| \(20t-5t^2\) | \(=\) | \(0\) |
| \(5t(4-t)\) | \(=\) | \(0\) |
\(t=0\) is the start, so it lands at \(t=\boldsymbol{4}\) s.
(ii) The path is symmetric, so the top is halfway between \(t=0\) and \(t=4\): at \(t=2\), \(h=40-20=\boldsymbol{20}\) m. Check: \(20t-5t^2=20\) gives \(5t^2-20t+20=0\), with \(\Delta=400-400=0\): the height \(20\) m is reached only once.
(iii) Set \(h=15\) and divide by \(5\).
| \(5t^2-20t+15\) | \(=\) | \(0\) |
| \(t^2-4t+3\) | \(=\) | \(0\) |
| \((t-1)(t-3)\) | \(=\) | \(0\) |
It is \(15\) m high at \(t=\boldsymbol{1}\) s (going up) and \(t=\boldsymbol{3}\) s (coming down). Both answers are valid.
(iv) The greatest height is reached only once, so \(kt-5t^2=5\) has exactly one solution.
| \(5t^2-kt+5\) | \(=\) | \(0\) |
| \(\Delta=k^2-100\) | \(=\) | \(0\) |
| \(k\) | \(=\) | \(\boldsymbol{10}\quad (k\gt 0)\) |
Check: \(h=10t-5t^2\) lands at \(t=2\), and at \(t=1\) the height is \(10-5=5\) m.
(i) Write the daily income in terms of \(k\).
(ii) Find the prices that give a daily income of \(\$1080\).
(i) Price \(=(5+0.5k)\) dollars and coffees sold \(=200-10k\).
| \(I\) | \(=\) | \((5+0.5k)(200-10k)\) |
| \(\) | \(=\) | \(1000+50k-5k^2\) |
(ii) Set \(I=1080\) and divide by \(5\).
| \(5k^2-50k+80\) | \(=\) | \(0\) |
| \(k^2-10k+16\) | \(=\) | \(0\) |
| \((k-2)(k-8)\) | \(=\) | \(0\) |
\(k=2\): price \(\boldsymbol{\$6.00}\), \(180\) coffees. \(k=8\): price \(\boldsymbol{\$9.00}\), \(120\) coffees. Both give \(\$1080\), so both prices are answers.
Common pitfalls
Frequently asked questions
How do you set up a quadratic word problem?
Let \(x\) stand for one unknown and write every other amount in terms of \(x\). Then use the fact given in the question, such as an area or a total, to write an equation. It is quadratic when two expressions in \(x\) are multiplied.
Why do quadratic word problems often have only one answer?
The equation has two solutions, but the situation may rule one out, for example a negative length, a negative time or a fraction of a person. State the reason when you reject a solution.
Can both solutions be correct?
Yes. Two different ticket prices can bring in the same income, and a thrown ball is at the same height twice, once going up and once coming down. Check each solution in the situation.
How do you show something never happens?
Form the quadratic equation for it and work out the discriminant \(b^2-4ac\). If it is negative, the equation has no real solutions, so the situation never occurs.
How do you find the greatest height of a thrown ball?
The ball is at each height twice, once going up and once coming down, except at the top. The top is halfway between the two times when \(h=0\). You can also set \(h=H\) and make the discriminant \(0\), because the greatest height is reached only once.