Changing the Subject: Squares and Roots

How to change the subject of a formula when the subject is squared, cubed or under a root: the order in which to undo the steps, when to write ±, and when to take only the positive root.

Before you start

Lesson 4.6: to change the subject, do the same to both sides, undoing what was done to the subject in reverse order.

Every positive number has two square roots: $3^2 = 9$ and ${(-3)}^2 = 9$ . The sign $\sqrt{9}$ means the positive one, $3$ . So if $x^2 = 9$ , then $x = 3$ or $x = -3$ , written $x = \pm 3$ ("plus or minus $3$ "). A negative number has no square root.

A number has only one cube root: $\sqrt[3]{8} = 2$ and $\sqrt[3]{-8} = -2$ .

Undoing a Square

Done to the subject	Undo it by
add $a$	subtract $a$
multiply by $a$	divide by $a$
square	take the square root, with $\pm$
take the square root	square
cube	take the cube root
take the cube root	cube

Square roots and $\pm$

If $x^2 = k$ (with $k$ positive), then $x = \pm\sqrt{k}$ .

When you take the square root of both sides, write $\pm$ , unless the letter is known to be positive (a length, a time, a speed, a mass). Then say so, and take only the positive root.

Squaring both sides, or taking a cube root, needs no $\pm$ .

Example 1: The Subject Squared

Make $u$ the subject of $v^2 = u^2 + 2as$ .

Solution

From $u$ : square it, then add $2as$ . Undo the adding first. Subtract $2as$ from both sides: $u^2 = v^2 - 2as$ .

Then undo the squaring. Take the square root of both sides: $u = \pm\sqrt{v^2 - 2as}$ .

Both signs work. Check with $a = 2$ , $s = 4$ : if $u = 3$ or $u = -3$ , then $u^2 = 9$ and $v^2 = 9 + 16 = 25$ . The new formula gives $\pm\sqrt{25 - 16} = \pm 3$ : both of them.

$u = \pm\sqrt{v^2 - 2as}$

Example 2: A Squared Bracket

Make $x$ the subject of $y = {(x - p)}^2 + q$ .

Solution

From $x$ : subtract $p$ , square, then add $q$ . Undo in reverse.

Subtract $q$ : $y - q = {(x - p)}^2$ .

Take the square root: $x - p = \pm\sqrt{y - q}$ .

Add $p$ : $x = p \pm \sqrt{y - q}$ .

Check with $p = 1$ , $q = 1$ , $y = 5$ : $x = 1 \pm 2$ , so $x = 3$ or $x = -1$ . Both give ${(x - 1)}^2 + 1 = 4 + 1 = 5\text{.}$

$x = p \pm \sqrt{y - q}$

Example 3: A Positive Quantity

(a) The area of a circle of radius $r$ is $A = \pi r^2$ . Make $r$ the subject. (b) An object of mass $m$ moving at speed $v$ has kinetic energy $E = \dfrac{1}{2}mv^2$ . Make $v$ the subject.

Solution

(a) Divide by $\pi$ : $r^2 = \dfrac{A}{\pi}$ . A radius is a length, so it is positive: $r = \sqrt{\dfrac{A}{\pi}}$ .

(b) Multiply by $2$ : $2E = mv^2$ . Divide by $m$ (a mass is not zero): $v^2 = \dfrac{2E}{m}$ . A speed is positive: $v = \sqrt{\dfrac{2E}{m}}$ .

Check (a) with $r = 3$ : $A = 9\pi$ , and $\sqrt{\dfrac{9\pi}{\pi}} = \sqrt{9} = 3\text{.}$

(a) $r = \sqrt{\dfrac{A}{\pi}}$ (b) $v = \sqrt{\dfrac{2E}{m}}$ (both positive, as a length and a speed)

Example 4: The Subject Squared in a Denominator

Make $x$ the subject of $y = \dfrac{k}{x^2}$ .

Solution

Multiply both sides by $x^2$ : $x^2 y = k$ .

Divide by $y$ ( $y \neq 0$ ): $x^2 = \dfrac{k}{y}$ .

Take the square root: $x = \pm\sqrt{\dfrac{k}{y}}$ .

Check with $k = 12$ , $x = -2$ : $y = \dfrac{12}{4} = 3$ , and $\pm\sqrt{\dfrac{12}{3}} = \pm 2$ , which includes $-2$ .

$x = \pm\sqrt{\dfrac{k}{y}}$

Undoing a Square Root

If the subject is under a square root, square both sides when the root is the last step left to undo. Squaring gives one answer, so no $\pm$ is needed.

Example 5: The Subject under a Square Root

Make the letter in brackets the subject. (a) $y = \sqrt{x + 3}$ $(x)$ (b) $t = 2\pi\sqrt{\dfrac{l}{g}}$ $(l)$ , the time of one swing of a pendulum of length $l$ .

Solution

(a) The square root was done last, so undo it first. Square both sides: $y^2 = x + 3$ . Subtract $3$ : $x = y^2 - 3$ .

(b) Divide by $2\pi$ : $\dfrac{t}{2\pi} = \sqrt{\dfrac{l}{g}}$ . Square both sides: $\dfrac{t^2}{4\pi^2} = \dfrac{l}{g}$ . Multiply by $g$ : $l = \dfrac{gt^2}{4\pi^2}$ .

Squaring needs no $\pm$ .

Check (b) with $g = 10$ , $l = 2.5$ : $t = 2\pi\sqrt{0.25} = \pi$ , and $\dfrac{10\pi^2}{4\pi^2} = 2.5$ .

(a) $x = y^2 - 3$ (b) $l = \dfrac{gt^2}{4\pi^2}$

Example 6: Squaring, Then a Square Root

Make $x$ the subject of $y = \sqrt{x^2 + 9}$ .

Solution

From $x$ : square it, add $9$ , then take the square root. Undo in reverse.

The root was done last. Square both sides: $y^2 = x^2 + 9$ .

Subtract $9$ : $x^2 = y^2 - 9$ .

Now undo the square: $x = \pm\sqrt{y^2 - 9}$ . Squaring the root away needed no $\pm$ ; taking the root of $x^2$ does.

Do not take the root term by term: $\sqrt{y^2 - 9}$ is not $y - 3$ .

Check with $x = 4$ and $x = -4$ : both give $y = \sqrt{16 + 9} = 5$ , and $\pm\sqrt{25 - 9} = \pm 4$ .

$x = \pm\sqrt{y^2 - 9}$

Cubes and Cube Roots

A cube is undone by taking the cube root, and a cube root by cubing. A number has only one cube root, so neither step needs $\pm$ .

Example 7: A Cube and a Cube Root

(a) The volume of a sphere of radius $r$ is $V = \dfrac{4}{3}\pi r^3$ . Make $r$ the subject. (b) Make $x$ the subject of $y = \sqrt[3]{2x + 1}$ .

Solution

(a) Multiply by $3$ : $3V = 4\pi r^3$ . Divide by $4\pi$ : $r^3 = \dfrac{3V}{4\pi}$ .

Take the cube root. A number has only one cube root, so there is no $\pm$ : $r = \sqrt[3]{\dfrac{3V}{4\pi}}$ .

(b) The cube root was done last. Cube both sides: $y^3 = 2x + 1$ . Subtract $1$ and divide by $2$ : $x = \dfrac{y^3 - 1}{2}$ .

Check (a) with $r = 3$ : $V = \dfrac{4}{3}\pi \times 27 = 36\pi\text{,}$ and $\sqrt[3]{\dfrac{108\pi}{4\pi}} = \sqrt[3]{27} = 3\text{.}$

Check (b) with $x = 13$ : $y = \sqrt[3]{27} = 3$ , and $\dfrac{27 - 1}{2} = 13$ .

(a) $r = \sqrt[3]{\dfrac{3V}{4\pi}}$ (b) $x = \dfrac{y^3 - 1}{2}$

Common Mistakes

Leaving out $\pm$ : from $u^2 = v^2 - 2as$ , $u = \pm\sqrt{v^2 - 2as}$ .
Writing $\pm$ for a letter that must be positive: a radius is $r = \sqrt{\dfrac{A}{\pi}}$ , not $\pm\sqrt{\dfrac{A}{\pi}}$ . Say why you take only the positive root.
Taking the square root term by term: $\sqrt{v^2 - 2as}$ is not $v - \sqrt{2as}$ . Check $v = 5$ , $2as = 16$ : $\sqrt{9} = 3$ , but $5 - 4 = 1$ .
Taking the root before the other terms have moved: from $y = x^2 + 4$ , subtract $4$ first, then $x = \pm\sqrt{y - 4}$ .
Writing $\pm$ after squaring or after a cube root: neither needs it.

Practice

Make the letter in brackets the subject. Write $\pm$ where both signs are possible. Check each answer with numbers.

Question 1

(a) $A = s^2$ , where $s$ is the side of a square $(s)$ (b) $y = x^2 - 5$ $(x)$ (c) $p = 3q^2$ $(q)$

ShowHide solution

(a) $s$ is a length, so positive: $s = \sqrt{A}$ .

(b) Add $5$ : $x^2 = y + 5$ , so $x = \pm\sqrt{y + 5}$ .

(a) $s = \sqrt{A}$ (b) $x = \pm\sqrt{y + 5}$ (c) $q = \pm\sqrt{\dfrac{p}{3}}$

Question 2

Make each letter the subject of $v^2 = u^2 + 2as$ in turn: (a) $a$ (b) $s$ (c) $v$ .

ShowHide solution

(a) $2as = v^2 - u^2$ , so $a = \dfrac{v^2 - u^2}{2s}$ . No root is needed: $a$ is not squared.

(b) Likewise $s = \dfrac{v^2 - u^2}{2a}$ .

Check (a) with $u = 1$ , $s = 2$ , $a = 3$ : $v^2 = 1 + 12 = 13$ , and $\dfrac{13 - 1}{4} = 3$ .

(a) $a = \dfrac{v^2 - u^2}{2s}$ (b) $s = \dfrac{v^2 - u^2}{2a}$ (c) $v = \pm\sqrt{u^2 + 2as}$

Question 3

(a) $y = {(x + 2)}^2 - 1$ $(x)$ (b) $y = 2{(x - 1)}^2$ $(x)$

ShowHide solution

(a) Add $1$ : $y + 1 = {(x + 2)}^2$ . Square root: $x + 2 = \pm\sqrt{y + 1}$ . Subtract $2$ : $x = -2 \pm \sqrt{y + 1}$ .

(b) Divide by $2$ : $\dfrac{y}{2} = {(x - 1)}^2$ . Square root: $x - 1 = \pm\sqrt{\dfrac{y}{2}}$ . Add $1$ : $x = 1 \pm \sqrt{\dfrac{y}{2}}$ .

Check (a) with $y = 8$ : $x = -2 \pm 3$ , so $1$ or $-5$ ; ${(1 + 2)}^2 - 1 = 8$ and ${(-5 + 2)}^2 - 1 = 8$ .

(a) $x = -2 \pm \sqrt{y + 1}$ (b) $x = 1 \pm \sqrt{\dfrac{y}{2}}$

Question 4

(a) $y = \sqrt{x - 2}$ $(x)$ (b) $y = 3\sqrt{x}$ $(x)$ (c) $t = 2\pi\sqrt{\dfrac{l}{g}}$ $(g)$

ShowHide solution

(a) Square: $y^2 = x - 2$ , so $x = y^2 + 2$ .

(b) Divide by $3$ , then square: $x = \dfrac{y^2}{9}$ .

(a) $x = y^2 + 2$ (b) $x = \dfrac{y^2}{9}$ (c) $g = \dfrac{4\pi^2 l}{t^2}$

Question 5

(a) The brightness of a lamp at a distance $d$ is $I = \dfrac{k}{d^2}$ $(d)$ (b) $y = \dfrac{1}{x^2 + 1}$ $(x)$

ShowHide solution

(a) Multiply by $d^2$ and divide by $I$ (a lamp that is on has $I \neq 0$ ): $d^2 = \dfrac{k}{I}$ . A distance is positive: $d = \sqrt{\dfrac{k}{I}}$ .

(b) $y{(x^2 + 1)} = 1$ . Divide by $y$ , which is never $0$ here ( $1$ divided by a number is never $0$ ): $x^2 + 1 = \dfrac{1}{y}$ and $x^2 = \dfrac{1}{y} - 1 = \dfrac{1 - y}{y}\text{.}$ So $x = \pm\sqrt{\dfrac{1 - y}{y}}$ .

Check (b) with $x = 2$ : $y = \dfrac{1}{5}$ , and $\dfrac{1 - \frac{1}{5}}{\frac{1}{5}} = 4$ , whose square roots are $\pm 2$ .

(a) $d = \sqrt{\dfrac{k}{I}}$ (b) $x = \pm\sqrt{\dfrac{1 - y}{y}}$

Question 6

(a) The volume of a cube of edge $s$ is $V = s^3$ $(s)$ (b) A cone whose height equals its radius $r$ has volume $V = \dfrac{1}{3}\pi r^3$ $(r)$ (c) $a = \sqrt[3]{b - 5}$ $(b)$

ShowHide solution

(a) $s = \sqrt[3]{V}$ .

(b) $3V = \pi r^3$ , so $r^3 = \dfrac{3V}{\pi}$ and $r = \sqrt[3]{\dfrac{3V}{\pi}}$ .

(c) Cube both sides: $a^3 = b - 5$ , so $b = a^3 + 5$ . Check with $b = -3$ : $a = \sqrt[3]{-8} = -2$ , and ${(-2)}^3 + 5 = -8 + 5 = -3\text{.}$

(a) $s = \sqrt[3]{V}$ (b) $r = \sqrt[3]{\dfrac{3V}{\pi}}$ (c) $b = a^3 + 5$

Question 7

(a) $p = \sqrt{q^2 - 4}$ $(q)$ (b) $r = \sqrt{\dfrac{3V}{\pi h}}$ $(V)$

ShowHide solution

(a) Square both sides: $p^2 = q^2 - 4$ . Add $4$ : $q^2 = p^2 + 4$ . Take the square root: $q = \pm\sqrt{p^2 + 4}$ .

Check (a) with $q = 2.5$ : $p = \sqrt{6.25 - 4} = \sqrt{2.25} = 1.5\text{,}$ and $\pm\sqrt{2.25 + 4} = \pm\sqrt{6.25} = \pm 2.5\text{.}$

(b) Square both sides: $r^2 = \dfrac{3V}{\pi h}$ . Multiply by $\pi h$ , then divide by $3$ : $V = \dfrac{\pi r^2 h}{3}$ . No root is taken, so no $\pm$ .

Check (b) with $r = 3$ , $h = 4$ : $V = \dfrac{36\pi}{3} = 12\pi$ , and $\sqrt{\dfrac{36\pi}{4\pi}} = \sqrt{9} = 3\text{.}$

(a) $q = \pm\sqrt{p^2 + 4}$ (b) $V = \dfrac{\pi r^2 h}{3}$

Question 8

A stone dropped from rest falls about $s = 5t^2$ metres in $t$ seconds. (a) Make $t$ the subject. (b) How long does it take to fall $45$ m?

ShowHide solution

(a) $t^2 = \dfrac{s}{5}$ . A time after the drop is positive: $t = \sqrt{\dfrac{s}{5}}$ .

(b) $t = \sqrt{9} = 3$ . Check: $5 \times 3^2 = 45$ .

(a) $t = \sqrt{\dfrac{s}{5}}$ (b) $3$ seconds

Question 9

A ladder $L$ m long leans against a vertical wall. Its foot is $d$ m from the wall on level ground, and its top is $h$ m up the wall, so that $L^2 = h^2 + d^2$ . (a) Make $h$ the subject. (b) A ladder $6.5$ m long has its foot $2.5$ m from the wall. How far up the wall does it reach?

ShowHide solution

(a) Subtract $d^2$ : $h^2 = L^2 - d^2$ . A height is positive: $h = \sqrt{L^2 - d^2}$ . It is not $L - d$ : the root of a difference is not the difference of the roots.

(b) $h = \sqrt{6.5^2 - 2.5^2} = \sqrt{42.25 - 6.25} = \sqrt{36} = 6\text{.}$

Check: $6^2 + 2.5^2 = 36 + 6.25 = 42.25 = 6.5^2\text{.}$

(a) $h = \sqrt{L^2 - d^2}$ (b) $6$ m

Question 10

The surface area of a ball of radius $r$ cm is $A = 4\pi r^2$ cm². (a) Make $r$ the subject. (b) A ball has a surface area of $1256$ cm². Taking $\pi = 3.14$ , find its radius.

ShowHide solution

(a) Divide by $4\pi$ : $r^2 = \dfrac{A}{4\pi}$ . A radius is a length, so positive: $r = \sqrt{\dfrac{A}{4\pi}}$ .

(b) $4\pi = 12.56$ , so $r^2 = \dfrac{1256}{12.56} = 100$ and $r = 10$ .

Check: $4 \times 3.14 \times 10^2 = 12.56 \times 100 = 1256\text{.}$

(a) $r = \sqrt{\dfrac{A}{4\pi}}$ (b) $10$ cm

Question 11

Find and correct the error. (a) Ravi makes $x$ the subject of $y = x^2 - 9$ and writes $x = \sqrt{y} + 3$ . (b) Ken makes $x$ the subject of $y = {(x + 1)}^2$ and writes $x = \pm\sqrt{y} + 1$ .

ShowHide solution

(a) Test $x = 4$ : $y = 16 - 9 = 7$ , but $\sqrt{7} + 3$ is about $5.6$ , not $4$ . He took the root term by term: $\sqrt{y + 9}$ is not $\sqrt{y} + 3$ . He also left out $\pm$ : $x = -4$ gives $y = 16 - 9 = 7$ too. Correct: add $9$ first, $x^2 = y + 9$ , so $x = \pm\sqrt{y + 9}$ ; with $y = 7$ this gives $\pm 4$ .

(b) Test $x = 2$ : $y = 9$ , but his formula gives $3 + 1 = 4$ or $-3 + 1 = -2$ , and neither is $2$ . From $x + 1 = \pm\sqrt{y}$ he added $1$ instead of subtracting it. Correct: $x = -1 \pm \sqrt{y}$ ; with $y = 9$ this gives $2$ or $-4$ , and ${(-4 + 1)}^2 = 9$ too.

(a) $x = \pm\sqrt{y + 9}$ (b) $x = -1 \pm \sqrt{y}$

Key Points

Undo the steps done to the subject in reverse order: the last one first.
Undo a square by taking the square root of both sides, and write $\pm$ : if $x^2 = k$ , then $x = \pm\sqrt{k}$ .
Take only the positive root when the letter must be positive (a length, a time, a speed, a mass), and say so.
Undo a square root by squaring both sides, a cube by taking the cube root, and a cube root by cubing; none of these needs $\pm$ .
Move the other terms away before taking a root: never take a root term by term.
Check by putting numbers into both forms, and with $\pm$ check both signs.

Changing the Subject: Squares and Roots

Before you start

Undoing a Square

Square roots and ±\pm±

Example 1: The Subject Squared

Example 2: A Squared Bracket

Example 3: A Positive Quantity

Example 4: The Subject Squared in a Denominator

Undoing a Square Root

Example 5: The Subject under a Square Root

Example 6: Squaring, Then a Square Root

Cubes and Cube Roots

Example 7: A Cube and a Cube Root

Common Mistakes

Practice

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Key Points

Square roots and $\pm$