E2.5 - Rearranging harder formulas (Extended) | Cambridge IGCSE International Mathematics

If a circle's area is known, rearranging $A=\pi r^2$ seems to give two possible values for $r$ . Which one can be a radius? This Extended E2.5 lesson rearranges powers and roots, collects a repeated subject, and uses the original formula and context to keep only valid values.

Powers and roots

A power or root changes the operation you must undo. First isolate the whole powered or rooted expression; only then apply the appropriate inverse. Do not take a square root of separate terms in a sum.

Principal square root

For $k\ge0$ , $\sqrt{k}$ is the non-negative number whose square is $k$ . The equation $x^2=k$ allows $x=\pm\sqrt{k}$ ; if $x\ge0$ is given, only the non-negative branch is permitted. At $k=0$ , the two signs give the same value, zero.

Thus $x^2=9$ allows $x=3$ and $x=-3$ , whereas $x=\sqrt9$ states $x=3$ . Squaring $y=\sqrt{x}$ gives $x=y^2$ , but the original restriction $y\ge0$ must be kept. Otherwise the squared equation would admit values the original rejects.

Worked example 1: isolating a square

Question. Make $x$ the subject of $y=3x^2+2$ , given $x\ge0$ .

Plan. Remove the addition and multiplier, then take the non-negative square root.

\begin{aligned} y-2&=3x^2 &&\text{subtract 2}\\ \frac{y-2}{3}&=x^2 &&\text{divide by 3}\\ x&=\sqrt{\frac{y-2}{3}} &&\text{use the given }x\ge0 \end{aligned}

Answer. $x=\sqrt{\frac{y-2}{3}}$ , with $y\ge2$ for a real value.

Check. Substitution gives $3\left(\sqrt{(y-2)/3}\right)^2+2=y$ . At $y=14$ , the answer is $x=2$ , and $3(2)^2+2=14$ .

Change the operation on $x$ from squaring to taking its principal square root. Squaring now undoes the root, but only after it has been isolated.

Worked example 2: isolating a square root

Question. Make $x$ the subject of $y=3\sqrt{x}+2$ .

Plan. Isolate $\sqrt{x}$ , record its non-negative value, then square the whole side.

\begin{aligned} y-2&=3\sqrt{x}\\ \frac{y-2}{3}&=\sqrt{x} &&\text{this side must be non-negative}\\ x&=\left(\frac{y-2}{3}\right)^2 &&\text{square both whole sides} \end{aligned}

Answer. $x=\left(\frac{y-2}{3}\right)^2$ , with $y\ge2$ .

Check. For $y\ge2$ , the square root of the proposed $x$ is $(y-2)/3$ , so $3\sqrt{x}+2=y$ . Numerically, $y=8$ gives $x=4$ , and $3\sqrt4+2=8$ . Taking $y=-4$ in the squared expression would fail the original, so the restriction is essential.

Odd powers behave differently: $x^3=k$ has the single real solution $x=\sqrt[3]{k}$ , including when $k<0$ . For an even power, use the sign information and the non-negative powered value before choosing a branch.

Watch out

$\sqrt{y-2}\ne\sqrt y-\sqrt2$ in general. Also, from $x^2=k$ , writing just $x=\sqrt{k}$ discards a negative solution unless a restriction justifies doing so.

Reverse the operations outside the root before squaring.

Repeated subject

When the subject appears in two terms, undoing the operations around only one occurrence leaves the other behind. Collect every term containing the subject on the same side, put all other terms on the other side, then extract the subject as a common factor.

The distributive law reverses this step: $ax+bx=x(a+b)$ . Once there is only one outside $x$ , divide by the whole factor in brackets, provided it is non-zero.

Worked example 3: two subject terms on one side

Question. Make $x$ the subject of $y=ax+bx$ , assuming $a+b\ne0$ .

Plan. Factorise out $x$ , then divide by its whole coefficient.

\begin{aligned} y&=x(a+b) &&\text{common factor }x\\ x&=\frac{y}{a+b} &&\text{divide by the non-zero factor} \end{aligned}

Answer. $x=\frac{y}{a+b}$ , with $a+b\ne0$ .

Check. Substitution gives $a\frac{y}{a+b}+b\frac{y}{a+b}=\frac{(a+b)y}{a+b}=y$ . For $a=2$ , $b=3$ , $y=20$ , the result $x=4$ gives $8+12=20$ .

Now the second subject term is on the other side. Collecting it first changes the sum of coefficients to a difference.

Worked example 4: subject terms on both sides

Question. Make $x$ the subject of $y+bx=ax$ , assuming $a\ne b$ .

Plan. Subtract $bx$ , factorise the remaining subject terms, then divide.

\begin{aligned} y&=ax-bx &&\text{subtract }bx\text{ from both sides}\\ y&=x(a-b) &&\text{factorise }x\\ x&=\frac{y}{a-b} &&\text{since }a-b\ne0 \end{aligned}

Answer. $x=\frac{y}{a-b}$ , with $a\ne b$ .

Check. In the original, the left side becomes $y+\frac{by}{a-b}=\frac{ay}{a-b}$ , which equals $ax$ . Numerically, $a=5$ , $b=2$ , $y=12$ gives $x=4$ , and $12+2(4)=5(4)=20$ .

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E2.5 - Rearranging harder formulas (Extended)

Powers and roots

Repeated subject

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