Solving Quadratic Equations Involving Squares

How to solve $x^2 = 49$ , $4x^2 = 25$ , $(x - 1)^2 = 9$ and $x^2 - 6x + 9 = 0$ by factorising with the difference of two squares and perfect squares, and why taking the square root gives two answers.

Before you start

In chapter 3 you met three identities: $a^2 - b^2 = (a + b)(a - b)\text{,}$ $(a + b)^2 = a^2 + 2ab + b^2$ and $(a - b)^2 = a^2 - 2ab + b^2\text{.}$ For example, $x^2 - 25 = (x + 5)(x - 5)$ and $x^2 - 8x + 16 = (x - 4)^2\text{.}$
In lesson 5.1 you solved a quadratic equation by making one side $0$ , factorising, and setting each factor equal to $0$ .

Using the Difference of Two Squares

The equation $x^2 - 9 = 0$ has no $x$ term. Its left side is $x^2 - 3^2$ , a difference of two squares, so it factorises at once.

Example 1: A difference of two squares

Solve $x^2 - 9 = 0$ .

Solution

$x^2 - 9 = x^2 - 3^2 = (x + 3)(x - 3)\text{,}$ so $(x + 3)(x - 3) = 0$ .

Either $x + 3 = 0$ , so $x = -3$ , or $x - 3 = 0$ , so $x = 3$ .

Check: $(-3)^2 - 9 = 9 - 9 = 0$ ✓ and $3^2 - 9 = 0$ ✓

$x = 3$ or $x = -3$ , which is written $x = \pm 3$ (read “plus or minus 3”).

The two roots are a number and its negative. That is no accident: $3^2 = 9$ and $(-3)^2 = 9$ as well, because a negative number times a negative number is positive. So the equation $x^2 = 9$ asks “which numbers square to $9$ ?”, and there are two of them.

Taking the square root of both sides

If $x^2 = n^2$ , then $x = n$ or $x = -n$ , written $x = \pm n$ .

This gives the same two roots as factorising, because $x^2 - n^2 = (x + n)(x - n)\text{.}$ It is a quicker way to write the same working.

In this lesson the number on the right is always a perfect square, such as $49 = 7^2$ , or the square of a fraction, such as $\frac{25}{4} = \left(\frac{5}{2}\right)^2\text{,}$ so the roots are whole numbers or fractions.

Example 2: Two ways

Solve $x^2 = 49$
(a) by factorising
(b) by taking the square root of both sides.

Solution

(a) Subtract $49$ from both sides: $x^2 - 49 = 0$ . Since $49 = 7^2$ , $(x + 7)(x - 7) = 0$ . So $x = -7$ or $x = 7$ .

(b) $49 = 7^2$ , so $x = \pm 7$ .

Check: $7^2 = 49$ ✓ and $(-7)^2 = 49$ ✓

$x = \pm 7$ by either way

A Number in Front of x²

When $x^2$ has a coefficient, as in $4x^2 = 64$ , divide both sides by it first, or take it out as a common factor. Either way you reach a difference of two squares.

Example 3: Divide first

Solve
(a) $4x^2 = 64$
(b) $3x^2 - 12 = 0$ .

Solution

(a) Divide both sides by $4$ : $x^2 = 16$ . Since $16 = 4^2$ , $x = \pm 4$ .

Check: $4 \times 4^2 = 4 \times 16 = 64$ ✓, and $(-4)^2$ is also $16$ ✓.

(b) Take out the common factor $3$ : $3(x^2 - 4) = 0$ , so $3(x + 2)(x - 2) = 0$ .

The factor $3$ is never $0$ , so $x + 2 = 0$ or $x - 2 = 0$ : $x = \pm 2$ .

Check: $3 \times 4 - 12 = 0$ ✓

(a) $x = \pm 4$
(b) $x = \pm 2$

Example 4: Roots that are fractions

Solve
(a) $4x^2 = 25$
(b) $9x^2 - 4 = 0$
(c) $x^2 = \frac{1}{4}$ .

Solution

(a) Divide by $4$ : $x^2 = \frac{25}{4}$ . Since $\frac{25}{4} = \left(\frac{5}{2}\right)^2\text{,}$ $x = \pm \frac{5}{2}$ .

By factorising instead: $4x^2 - 25 = (2x)^2 - 5^2 = (2x + 5)(2x - 5) = 0\text{,}$ so $x = -\frac{5}{2}$ or $x = \frac{5}{2}$ , the same roots.

(b) $9x^2 - 4 = (3x)^2 - 2^2 = (3x + 2)(3x - 2) = 0\text{.}$ Either $3x + 2 = 0$ , so $x = -\frac{2}{3}$ , or $3x - 2 = 0$ , so $x = \frac{2}{3}$ .

Check (b): $9 \times \frac{4}{9} - 4 = 4 - 4 = 0$ ✓

(c) $\frac{1}{4} = \left(\frac{1}{2}\right)^2\text{,}$ so $x = \pm \frac{1}{2}$ . Check: $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ ✓

(a) $x = \pm \frac{5}{2}$
(b) $x = \pm \frac{2}{3}$
(c) $x = \pm \frac{1}{2}$

When x² Is 0 or a Negative Number

A positive number times itself is positive; a negative number times itself is also positive; and $0 \times 0 = 0$ . So the square of a number is never negative.

Solving x² = a number

If the number is positive, such as $x^2 = 36$ : two roots, $x = \pm 6$ .

If the number is $0$ : $x^2 = 0$ means $x \times x = 0$ , so $x = 0$ , and the equation has two equal roots.

If the number is negative, such as $x^2 = -16$ : no number squared gives a negative number, so the equation has no solution.

Example 5: Zero and negative

Solve, if possible,
(a) $x^2 + 16 = 0$
(b) $5x^2 = 0$ .

Solution

(a) Subtract $16$ : $x^2 = -16$ . No number squared is negative, so there is no value of $x$ that makes this true.

(b) Divide by $5$ : $x^2 = 0$ , so $x = 0$ .

(a) There is no solution.
(b) $x = 0$ (two equal roots).

A Squared Bracket

In $(x - 1)^2 = 9$ , the whole bracket $x - 1$ is squared. Treat the bracket as one quantity: it squares to $9$ , so it is $3$ or $-3$ .

Example 6: Two ways again

Solve $(x - 1)^2 = 9$ .

Solution

By square roots. $x - 1 = \pm 3$ . Either $x - 1 = 3$ , so $x = 4$ , or $x - 1 = -3$ , so $x = -2$ .

By factorising. Subtract $9$ : $(x - 1)^2 - 3^2 = 0$ . This is a difference of two squares, $A^2 - B^2 = (A + B)(A - B)\text{,}$ with $A = x - 1$ and $B = 3$ :

$\big((x - 1) + 3\big)\big((x - 1) - 3\big) = 0\text{,}$ which is $(x + 2)(x - 4) = 0$ . So $x = -2$ or $x = 4$ .

Check $x = 4$ : $(4 - 1)^2 = 9$ ✓. Check $x = -2$ : $(-2 - 1)^2 = (-3)^2 = 9$ ✓

$x = 4$ or $x = -2$

Example 7: A coefficient inside the bracket

Solve $(2x + 1)^2 = 25$ .

Solution

$2x + 1 = \pm 5$ .

Either $2x + 1 = 5$ , so $2x = 4$ and $x = 2$ , or $2x + 1 = -5$ , so $2x = -6$ and $x = -3$ .

Check $x = 2$ : $5^2 = 25$ ✓. Check $x = -3$ : $(-6 + 1)^2 = (-5)^2 = 25$ ✓

$x = 2$ or $x = -3$

Notice that the two roots here are not a number and its negative. The $\pm$ belongs to the bracket, $2x + 1$ , not to $x$ .

Perfect Squares: Two Equal Roots

Some quadratic expressions are perfect squares: $x^2 - 6x + 9 = (x - 3)^2\text{.}$ You can spot one when the first and last terms are squares ( $x^2$ and $9 = 3^2$ ) and the middle term, ignoring its sign, is twice their product: $2 \times x \times 3 = 6x$ .

An equation “perfect square $= 0$ ” has two equal roots, as $(x - 2)^2 = 0$ did in lesson 5.1.

Example 8: Spot the perfect square

Solve
(a) $x^2 - 6x + 9 = 0$
(b) $x^2 + 10x + 25 = 0$
(c) $4x^2 + 4x + 1 = 0$ .

Solution

(a) $x^2 - 6x + 9 = (x - 3)^2\text{,}$ so $(x - 3)^2 = 0$ and $x = 3$ . Check: $9 - 18 + 9 = 0$ ✓

(b) $25 = 5^2$ and $10x = 2 \times x \times 5$ : $(x + 5)^2 = 0$ , so $x = -5$ . Check: $25 - 50 + 25 = 0$ ✓

(c) $4x^2 = (2x)^2$ , $1 = 1^2$ and $4x = 2 \times 2x \times 1$ : $(2x + 1)^2 = 0$ . So $2x + 1 = 0$ and $x = -\frac{1}{2}$ .

Check (c): $4 \times \frac{1}{4} + 4 \times (-\frac{1}{2}) + 1 = 1 - 2 + 1 = 0$ ✓

(a) $x = 3$
(b) $x = -5$
(c) $x = -\frac{1}{2}$ ; each equation has two equal roots.

Choosing the Quicker Way

Take the square root of both sides only when one side is a square (of $x$ or of a bracket) and the other side is a number. If there is an $x$ on the other side too, rearrange so that one side is $0$ and factorise, as in lesson 5.2.

Example 9: Which way?

Solve
(a) $x^2 = 36$
(b) $x^2 = 6x$ .

Solution

(a) One side is $x^2$ and the other a number: take the square root. $x = \pm 6$ .

(b) There is an $x$ on both sides. Taking the square root does not help, and dividing by $x$ would lose $x = 0$ . Subtract $6x$ : $x^2 - 6x = 0$ , so $x(x - 6) = 0$ , and $x = 0$ or $x = 6$ .

Check (b): $x = 0$ gives $0 = 0$ ✓; $x = 6$ gives $36 = 36$ ✓

(a) $x = \pm 6$
(b) $x = 0$ or $x = 6$

Common Mistakes

Giving only the positive root: $x^2 = 25$ gives $x = 5$ or $x = -5$ , since $(-5)^2 = 25$ .
Losing the negative case for a bracket: $(x - 1)^2 = 9$ gives $x - 1 = 3$ or $x - 1 = -3$ , so $x = 4$ or $x = -2$ .
Writing an answer for $x^2 = -16$ . No number squared is negative: the equation has no solution.
Confusing $x^2 - 9$ with $(x - 3)^2$ . $x^2 - 9 = (x + 3)(x - 3)\text{,}$ but $(x - 3)^2 = x^2 - 6x + 9\text{.}$
Taking the square root before dividing: $4x^2 = 25$ gives $x^2 = \frac{25}{4}$ first, then $x = \pm \frac{5}{2}$ .
Taking the square root when there is an $x$ on the other side, as in $x^2 = 6x$ . Rearrange and factorise.

Practice Questions

Solve each equation, and check your roots. The questions get harder as you go.

Question 1

Solve $x^2 - 64 = 0$ by factorising.

ShowHide solution

$x^2 - 8^2 = (x + 8)(x - 8) = 0\text{,}$ so $x = -8$ or $x = 8$ .

Check: $64 - 64 = 0$ for both, since $(-8)^2 = 64$ ✓

$x = \pm 8$

Question 2

Solve $x^2 = 81$ .

ShowHide solution

$81 = 9^2$ , so $x = \pm 9$ . (By factorising: $(x + 9)(x - 9) = 0$ .)

Check: $9^2 = 81$ and $(-9)^2 = 81$ ✓

$x = \pm 9$

Question 3

Solve $2x^2 = 50$ .

ShowHide solution

Divide by $2$ : $x^2 = 25$ , so $x = \pm 5$ .

Check: $2 \times 25 = 50$ ✓

$x = \pm 5$

Question 4

Solve $4x^2 - 400 = 0$ .

ShowHide solution

Take out $4$ : $4(x^2 - 100) = 0$ , so $4(x + 10)(x - 10) = 0\text{,}$ and $x = \pm 10$ .

Check: $4 \times 100 - 400 = 0$ ✓

$x = \pm 10$

Question 5

Solve $25x^2 = 9$ .

ShowHide solution

Divide by $25$ : $x^2 = \frac{9}{25} = \left(\frac{3}{5}\right)^2\text{,}$ so $x = \pm \frac{3}{5}$ .

Check: $25 \times \frac{9}{25} = 9$ ✓

$x = \pm \frac{3}{5}$

Question 6

Solve $16x^2 - 1 = 0$ .

ShowHide solution

$(4x)^2 - 1^2 = (4x + 1)(4x - 1) = 0\text{.}$ Either $4x + 1 = 0$ , so $x = -\frac{1}{4}$ , or $4x - 1 = 0$ , so $x = \frac{1}{4}$ .

Check: $16 \times \frac{1}{16} - 1 = 0$ ✓

$x = \pm \frac{1}{4}$

Question 7

Solve, if possible,
(a) $x^2 + 25 = 0$
(b) $3x^2 = 0$
(c) $x^2 - 1 = 0$ .

ShowHide solution

(a) $x^2 = -25$ . No number squared is negative, so there is no solution.

(b) $x^2 = 0$ , so $x = 0$ (two equal roots).

(a) no solution
(b) $x = 0$
(c) $x = \pm 1$

Question 8

Solve $(x + 2)^2 = 16$ .

ShowHide solution

$x + 2 = \pm 4$ . Either $x + 2 = 4$ , so $x = 2$ , or $x + 2 = -4$ , so $x = -6$ .

Check $x = 2$ : $4^2 = 16$ ✓. Check $x = -6$ : $(-4)^2 = 16$ ✓

$x = 2$ or $x = -6$

Question 9

Solve $(3x - 1)^2 = 49$ .

ShowHide solution

$3x - 1 = \pm 7$ .

Either $3x - 1 = 7$ , so $3x = 8$ and $x = \frac{8}{3}$ , or $3x - 1 = -7$ , so $3x = -6$ and $x = -2$ .

Check $x = \frac{8}{3}$ : $(8 - 1)^2 = 49$ ✓. Check $x = -2$ : $(-6 - 1)^2 = 49$ ✓

$x = \frac{8}{3}$ or $x = -2$

Question 10

Solve
(a) $x^2 - 14x + 49 = 0$
(b) $9x^2 - 12x + 4 = 0$ .

ShowHide solution

(a) $49 = 7^2$ and $14x = 2 \times x \times 7$ : $(x - 7)^2 = 0$ , so $x = 7$ . Check: $49 - 98 + 49 = 0$ ✓

(b) $9x^2 = (3x)^2$ , $4 = 2^2$ and $12x = 2 \times 3x \times 2$ : $(3x - 2)^2 = 0$ , so $x = \frac{2}{3}$ .

Check (b): $9 \times \frac{4}{9} - 12 \times \frac{2}{3} + 4 = 4 - 8 + 4 = 0$ ✓

(a) $x = 7$
(b) $x = \frac{2}{3}$ ; each has two equal roots.

Question 11

Daniel solved $(x + 3)^2 = 25$ like this: “ $x + 3 = 5$ , so $x = 2$ .” What has he missed? Give both roots.

ShowHide solution

The bracket squares to $25$ , so it could be $5$ or $-5$ . He has missed $x + 3 = -5$ .

$x + 3 = -5$ gives $x = -8$ . Check: $(-8 + 3)^2 = (-5)^2 = 25$ ✓

$x = 2$ or $x = -8$

Question 12

Choose the quicker way for each, and solve.
(a) $x^2 = 144$
(b) $x^2 = 12x$
(c) $x^2 - 12x + 36 = 0$

ShowHide solution

(a) A square equal to a number: $144 = 12^2$ , so $x = \pm 12$ .

(b) An $x$ on both sides: $x^2 - 12x = 0$ , so $x(x - 12) = 0$ , and $x = 0$ or $x = 12$ .

(a) $x = \pm 12$
(b) $x = 0$ or $x = 12$
(c) $x = 6$

Key Points

$x^2 - n^2 = 0$ factorises as $(x + n)(x - n) = 0$ , so $x = \pm n$ .
If $x^2 = n^2$ , then $x = \pm n$ : the square root of both sides gives the same two roots as factorising.
With a number in front of $x^2$ , divide by it (or take it out) first: $4x^2 = 25$ gives $x = \pm \frac{5}{2}$ .
$x^2 = 0$ has two equal roots, $x = 0$ ; $x^2 =$ a negative number has no solution.
For a squared bracket, the $\pm$ goes with the bracket: $(x - 1)^2 = 9$ gives $x - 1 = \pm 3$ , so $x = 4$ or $x = -2$ .
A perfect square equal to $0$ , such as $(x - 3)^2 = 0$ , has two equal roots.
If there is an $x$ on both sides, rearrange and factorise; never take the square root or divide by $x$ .