Equations

Constructing harder expressions Extended

Extended questions often need an expression with a product in it. Two consecutive whole numbers are $n$ and $n+1$. Two consecutive even numbers are $2n$ and $2n+2$: every even number is a multiple of 2, and the next even number is 2 more. Two consecutive odd numbers are $2n+1$ and $2n+3$.

So the product of two consecutive even numbers is $2n(2n+2)=4n^2+4n$. Notice that this is always a multiple of 4, which is the kind of fact an exam question may ask you to show.

Expressions like these lead to quadratic equations, which you will solve later in this lesson.

Fractional equations Extended

When the unknown is in a denominator, or there are several fractions, multiply every term on both sides by the lowest common denominator. That clears all the fractions at once and leaves an equation you already know how to solve.

• Numerical denominators. For $\frac{x+1}{2}-\frac{x-3}{5}=2$, multiply every term by 10: $5(x+1)-2(x-3)=20$. Watch the bracket after the minus sign: $-2(x-3)=-2x+6$. Then $5x+5-2x+6=20$, so $3x=9$ and $x=3$.
• An algebraic denominator. For $\frac{x}{3x-2}=2$, multiply both sides by $(3x-2)$: $x=6x-4$, so $5x=4$ and $x=\frac{4}{5}$.
• Two algebraic fractions. Multiply by the product of the denominators, as in Worked example 6. The result is often a quadratic equation.

Finally, check that no solution makes a denominator zero, because dividing by zero is impossible. Such a value must be rejected.

Quadratic equations by factorising Extended

A quadratic equation has an $x^2$ term and no higher power of $x$. Rearrange it so one side is 0, then factorise. The method rests on one fact: if two numbers multiply to give 0, at least one of them is 0. So if $(x-5)(x+3)=0$, either $x-5=0$ or $x+3=0$, giving $x=5$ or $x=-3$.

The zero is essential. $(x-5)(x+3)=9$ tells you nothing about either bracket on its own, so always rearrange to $=0$ first.

When the $x^2$ term has a coefficient, as in $2x^2+5x-12=0$, find two numbers that multiply to $2\times (-12)=-24$ and add to 5: they are 8 and $-3$. Split the middle term and factorise in pairs: $2x^2+8x-3x-12=2x(x+4)-3(x+4)=(2x-3)(x+4)$. So $x=\frac{3}{2}$ or $x=-4$.

Never divide both sides by $x$. In $x^2=7x$, dividing by $x$ loses the solution $x=0$. Instead write $x^2-7x=0$, so $x(x-7)=0$, and $x=0$ or $x=7$.

Completing the square Extended

Expanding $(x+p{)}^2$ gives $x^2+2px+p^2$, so the number in the bracket is half the coefficient of $x$. To write $x^2+bx+c$ in completed square form:

1. Halve the coefficient of $x$ and write the bracket $(x+\frac{b}{2}{)}^2$.
2. That bracket contains an extra $(\frac{b}{2}{)}^2$, so subtract it, then include $c$.

For $x^2-8x+5$: half of $-8$ is $-4$, and $(x-4{)}^2=x^2-8x+16$, which is 11 more than we want. So $x^2-8x+5=(x-4{)}^2-11$.

If the $x^2$ term has a coefficient, take it out of the $x$ terms first: $2x^2+12x+7=2(x^2+6x)+7=2((x+3{)}^2-9)+7=2(x+3{)}^2-11$.

Completed square form solves equations that do not factorise, and gives exact answers in surd form. If $(x-4{)}^2-11=0$ then $(x-4{)}^2=11$, so $x-4=\pm \sqrt{11}$ and $x=4\pm \sqrt{11}$. The $\pm$ is there because both $\sqrt{11}$ and $-\sqrt{11}$ square to give 11.

The quadratic formula Extended

For any quadratic equation $a{x}^2+bx+c=0$,
$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
The formula is printed in the List of formulas, so you do not need to memorise it, but you do need to use it accurately. It comes from completing the square on $a{x}^2+bx+c$ in general, which is why it works for every quadratic.

When a question asks for answers to 2 decimal places or 3 significant figures, the examiner is telling you the quadratic does not factorise: use the formula.

• Rearrange to $=0$ first, then read off $a$, $b$ and $c$ with their signs.
• Put negative values in brackets: if $b=-5$ then $b^2=(-5{)}^2=25$, not $-25$.
• Work out $b^2-4ac$ first, then the two answers separately, and round only at the end.
• If $b^2-4ac$ is negative there are no solutions, because no real number squares to a negative number.

Simultaneous equations: one linear, one non-linear Extended

When one equation is linear and the other has a squared term, such as $y=x^2-3x+1$ and $y=x-2$, elimination does not work. Use substitution: make $x$ or $y$ the subject of the linear equation and substitute it into the other one. The result is a quadratic in one letter, which usually has two solutions, so there are usually two pairs of answers. On a graph, a straight line can cross a curve at two points.

Pair each value of $x$ with its own value of $y$, and find $y$ from the linear equation: it is quicker and less error-prone.

Changing the subject: harder formulas Extended

Two harder situations appear at Extended level.

The subject appears twice. Collect every term containing the subject on one side and everything else on the other, then factorise so the subject appears once, and divide. For $p=\frac{3t+2}{t-1}$:

1. multiply both sides by $(t-1)$: $pt-p=3t+2$;
2. collect the $t$ terms on one side: $pt-3t=p+2$;
3. factorise: $t(p-3)=p+2$;
4. divide: $t=\frac{p+2}{p-3}$.

The subject is squared or under a root. Isolate the power or root, then undo it: a square is undone by a square root, and a square root by squaring. For $V=\frac{1}{3}\pi r^{2}h$: $3V=\pi r^{2}h$, so $r^2=\frac{3V}{\pi h}$ and $r=\sqrt{\frac{3V}{\pi h}}$. Here $r$ is a length, so only the positive root is wanted.