Equations and Inequalities: O-Level E-Math Notes

Three real situations, each modelled by an equation from this chapter. The pictures are drawn first, then the graph is placed on top by script so it matches the problem exactly.

Maths In Real Life 1

The fountain jet

Water from a nozzle follows a parabola because gravity pulls it down at a steady rate while it keeps moving forward.

At a water play area, a jet of water leaves a nozzle and its height is modelled by $h = -0.5x^2 + 2x + 0.5$, where $h$ metres is the height above the ground and $x$ metres is the horizontal distance from the nozzle.

• (a)Find the height of the nozzle.
• (b)Find how far from the nozzle the water lands, correct to $3$ significant figures.
• (c)Find the greatest height of the jet and the horizontal distance at which it occurs.

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(a)

$\displaystyle h = -0.5(0)^2 + 2(0) + 0.5$

Put $x = 0$: that is the nozzle.

$\displaystyle h = \frac{1}{2}$

The nozzle is $0.5$ m above the ground.

(b)

$\displaystyle -0.5x^2 + 2x + 0.5 = 0$

The jet lands when the height $h$ is $0$.

$\displaystyle x^2 - 4x - 1 = 0$

Multiply every term by $-2$ to clear the decimals and make $x^2$ positive.

$\displaystyle a = 1,\ b = -4,\ c = -1$

Compare with $ax^2 + bx + c = 0$ and copy each coefficient with its sign.

$\displaystyle x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$

The quadratic formula. It is NOT on the exam formula list, so memorise it.

$\displaystyle x = \dfrac{-(-4) \pm \sqrt{(-4)^{2} - 4(1)(-1)}}{2(1)}$

Substitute. Put brackets round every negative number.

$\displaystyle x = \dfrac{4 \pm \sqrt{20}}{2}$

Work out $b^2 - 4ac$ first: it is the number under the root.

$\displaystyle x = \dfrac{4 \pm 2\sqrt{5}}{2}$

Simplify the surd where you can: take out the square factor.

$\displaystyle x = 2 \pm \sqrt{5}$

Divide every term on top and the bottom by the same number: the bottom becomes $1$, so drop it.

$\displaystyle x = -0.236 \text{ or } x = 4.24$

Use your calculator for the $+$ and the $-$. Round to 3 significant figures only now, at the end.

The other root is $x \approx -0.236$. It is behind the nozzle, so it is rejected.

A distance forward cannot be negative.

$\displaystyle x = 4.24$

The water lands about this far from the nozzle, in metres.

(c)

$\displaystyle h = -0.5x^2 + 2x + 0.5 = -0.5(x^2 - 4x) + 0.5 = -0.5(x - 2)^2 + 2 + 0.5$

Take out $-0.5$, then complete the square inside: $x^2 - 4x = (x - 2)^2 - 4$.

$\displaystyle h = -0.5(x - 2)^2 + \frac{5}{2}$

In the form $h = a(x - p)^2 + k$ with $a < 0$: the turning point $(p, k)$ is a maximum.

A square is never negative, so $-0.5(x - 2)^2 \le 0$ and $h$ is greatest when $x = 2$.

Read the greatest height straight from the completed-square form.

$\displaystyle h = \frac{5}{2},\ x = 2$

The greatest height and the distance at which it happens.

Final answer

(a)The nozzle is $\frac{1}{2}$ m above the ground.

(b)The water lands $4.24$ m from the nozzle (the other root is negative, so it is rejected).

(c)The greatest height is $\frac{5}{2}$ m, at $x = 2$ m.

Maths In Real Life 2

Two kayak shops

A straight-line graph describes any cost that has a fixed fee plus a steady rate.

At East Coast Park, Shop A hires out a kayak for a fixed fee of $\char"24{}12$ plus $\char"24{}6$ per hour. Shop B charges $\char"24{}10$ per hour with no fixed fee.

• (a)Form an equation and find the number of hours for which the two shops charge the same.
• (b)For a hire of $2$ hours, which shop is cheaper? Explain using the graph.

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(a)

Let $x$ be the number of hours. Shop A charges $12 + 6x$ dollars and Shop B charges $10x$ dollars.

Fixed fee plus rate times hours.

$\displaystyle 12 + 6x = 10x$

The two costs are equal.

$\displaystyle 12 = 4x$

Subtract $6x$ from both sides.

$\displaystyle x = 3$

Divide by $4$.

$\displaystyle 12+6(3) = 30$

Check: both shops charge the same at this number of hours. Left side:

$\displaystyle 10(3) = 30$

Right side: the same value, so the root is correct.

(b)

$\displaystyle 12 + 6(2) = 24$

Shop A at $2$ hours: fixed fee plus rate times hours.

$\displaystyle 10(2) = 20$

Shop B at $2$ hours: rate times hours.

Shop B costs $\char"24{}20$ and Shop A costs $\char"24{}24$, so Shop B is cheaper. On the graph, Shop B's line is below Shop A's line until the lines cross.

Compare the two costs.

Final answer

(a)The shops charge the same at $3$ hours, when both charge $\char"24{}30$.

(b)Shop B is cheaper: $\char"24{}20$ against $\char"24{}24$.

Maths In Real Life 3

The community garden plot

Among all rectangles with the same perimeter, the square has the greatest area.

A community garden has a rectangular vegetable plot fenced by $40$ m of fencing. Its area is $96$ m².

• (a)Form an equation and find the possible dimensions of the plot.
• (b)Find the greatest possible area that $40$ m of fencing can enclose, and the width that gives it.

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(a)

The fence is $40$ m long, so length plus width is $20$ m. Let the width be $x$ m: then the length is $20 - x$ m.

Half the perimeter is length plus width.

$\displaystyle x(20-x)=96$

Area = width $\times$ length.

$\displaystyle x^{2} - 20x + 96 = 0$

Rearrange so that one side is $0$ and the $x^2$ term is positive.

$\displaystyle (x - 12)(x - 8) = 0$

Factorise (look for two numbers that fit).

$\displaystyle x - 12 = 0 \text{ or } x - 8 = 0$

If a product is $0$, one of the factors is $0$: set EACH factor equal to $0$.

$\displaystyle x = 8 \text{ or } x = 12$

Solve each little equation.

Both roots give a real plot: a $8$ m by $12$ m rectangle either way round.

Check each answer against the story.

(b)

$\displaystyle A = x(20 - x) = -(x - 10)^2 + 100$

Complete the square on $20x - x^2$: $-(x^2 - 20x) = -(x - 10)^2 + 100$.

A square is never negative, so $-(x - 10)^2 \le 0$ and $A$ is greatest when $x = 10$.

Read the maximum straight from the completed-square form.

$\displaystyle A_{\max} = 100 \text{ when } x = 10$

Then the length is $20 - 10 = 10$ as well: a square.

Final answer

(a)The plot is $8$ m by $12$ m.

(b)The greatest area is $100$ m², when the width is $10$ m (a square).

An equation says two expressions are equal, like a balance that is perfectly level. To solve it, find the value of the unknown that keeps the balance true. The one rule that never fails: do the same thing to both sides.

Linear equations

A linear equation has $x$ to the power $1$ only. Collect the $x$-terms on one side and the numbers on the other, then divide by the coefficient of $x$. Always finish by checking your answer in the original equation.

Fractional equations that reduce to linear equations

When the unknown sits inside fractions, clear the fractions first. Multiply every term on both sides by the lowest common multiple (LCM) of the denominators. After that it is an ordinary linear equation.

A quadratic equation has $x^2$ as its highest power. In standard form it is $ax^2 + bx + c = 0$, with $a \ne 0$. The values of $x$ that make it true are called its roots (or solutions). Unlike a linear equation, a quadratic can have two roots.

Quadratics come in a few shapes. Learn to spot them quickly.

Many quadratics refuse to factorise neatly. The quadratic formula solves every one of them, factorisable or not.

Completing the square rewrites $x^2 + px + q$ as $(x - h)^2 + k$. Then the unknown appears only once, so you can solve by taking a square root. K310 asks for it for quadratics of the form $y = x^2 + px + q$.