Applications of Quadratic Equations
Every question with a fully worked solution.
- Applications of Quadratic Equations - Video - Solving Quadratic Equations by Factoring (Word Problems) 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 9 maths page covers choosing a linear or quadratic model, falling objects and heights, number and formula problems, lengths and areas, Pythagoras, paths and ages, 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 solutions, the situation never happens.
Choose the model. Not every problem is quadratic. Simple interest, \(I = Prn\), grows by the same amount every year (\(\$2000\) at \(4\%\) p.a. earns \(\$80\) each year), so it is a linear model. A falling object, an area, or a price times a number sold, where both amounts change, gives a quadratic model. After solving, decide whether the answer is sensible and whether the model still fits.
Relationships that often produce quadratics:
| Situation | Relationship | In terms of \(x\) |
|---|---|---|
| Falling object | \(d=4.9t^2\) | divide by \(4.9\), then take the square root |
| 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 \(x+4\) |
| Rectangle with a path round it | \((l+2x)(w+2x)\) | \(x\) is the width of the path |
| Ages | now \(x\), in \(n\) years \(x+n\) | add \(n\) to every age |
| 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 product).
- Rearrange to \(ax^2+bx+c=0\), dividing by any common factor, then solve by factorising. Clear any fractions or decimals first. If there is no \(x\) term, take the square root.
- 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 \(h=25t-5t^2\), set \(h\) equal to the height asked for. For \(h=30\): \(5t^2-25t+30=0\), divide by \(5\): \(t^2-5t+6=0\), so \(t=2\) s (going up) or \(t=3\) s (coming down). The greatest height is halfway between the two times at ground level: \(h=5t(5-t)=0\) at \(t=0\) and \(t=5\), so the top is at \(t=2.5\) s, where \(h=31.25\) m.
- 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) A ball is dropped from a balcony \(30\) m above the ground. After \(t\) seconds it has fallen \(d=4.9t^2\) metres. How long does it take to reach the ground, correct to \(1\) decimal place?
(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) It reaches the ground when it has fallen \(30\) m. There is no \(t\) term, so get \(t^2\) on its own and take the square root.
| \(4.9t^2\) | \(=\) | \(30\) |
| \(t^2\) | \(=\) | \(\dfrac{30}{4.9}\approx 6.1224\) |
| \(t\) | \(=\) | \(\pm\sqrt{6.1224}\approx \pm 2.474\) |
A time cannot be negative, so it takes about \(\boldsymbol{2.5}\) seconds.
(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) A rectangular pool \(9\) m by \(5\) m is surrounded by a path of the same width, \(x\) m, all the way round. The pool and path together cover \(165\ \text{m}^2\). Find the width of the path.
(ii) Tom is \(4\) years older than his sister Ella. In \(3\) years, the product of their ages will be \(140\). How old are they now?
(i) The path adds \(x\) to each end, so the outside measures \((9+2x)\) m by \((5+2x)\) m.
| \((9+2x)(5+2x)\) | \(=\) | \(165\) |
| \(4x^2+28x+45\) | \(=\) | \(165\) |
| \(x^2+7x-30\) | \(=\) | \(0\) |
| \((x+10)(x-3)\) | \(=\) | \(0\) |
A width cannot be negative, so the path is \(\boldsymbol{3}\) m wide. Check: \(15\times 11=165\).
(ii) Let Ella be \(x\) now, so Tom is \(x+4\). In \(3\) years they will be \(x+3\) and \(x+7\).
| \((x+3)(x+7)\) | \(=\) | \(140\) |
| \(x^2+10x+21\) | \(=\) | \(140\) |
| \(x^2+10x-119\) | \(=\) | \(0\) |
| \((x+17)(x-7)\) | \(=\) | \(0\) |
An age cannot be negative, so Ella is \(\boldsymbol{7}\) and Tom is \(\boldsymbol{11}\). Check: \(10\times 14=140\).
(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.
What if the formula has decimals?
Multiply every term by \(10\), \(100\) or another number that clears the decimals. Then rearrange so one side is zero, factorise, and check the answer in the original formula.
How do you set up an age problem?
Let \(x\) be one person's age now and write the other ages in terms of \(x\). In \(n\) years every age increases by \(n\), so add \(n\) to each age before you multiply.