# Lesson 16: Completing the square for ax2+bx+c

Introduction

As with factorising to double brackets, completing the square also has to be able to deal with a coefficient of $$~x^2~$$ greater than 1. The completed square form is then $$~a(x+p)^2+q~$$.

In Section A, you'll get the opportunity to practise the method with nice numbers (I've made them artificially nice by ensuring $$~b~$$ is always a multiple of $$~2a~$$). In Section B, it's back to reality!

SECTION A

Let's say you're asked to write $$~2x^2+12x+7~$$ in the form $$~a(x+p)^2+q~$$.

Step 1

Factorise the first two terms to a single bracket by taking out a factor of $$~a$$. Here, $$~a~$$ is 2 so we take out a factor of 2.

$$~2[x^2+6x]+7~$$

Step 2

Now complete the square on the bit in the square brackets. That's $$~x^2+6x~$$. You should know how to do this from Lesson 15.

Half the coefficient of $$~x~$$ is $$~+3~$$

$$~(+3)^2=9~$$

The constant term is $$~0~$$. In other words, there isn't one!

So we have $$~(x+3)^2-9~$$

Step 3

Put this in the square brackets instead of $$~x^2+6x$$.

$$~2[(x+3)^2-9]+7~$$

Step 4

Now expand the square brackets back out by multiplying by $$~a~$$.

$$~2(x+3)^2-18+7~$$

Step 5

Simplify and relax!

$$~2(x+3)^2-11~$$

As always, to master this you will have to practise it! Check the worked solutions as you go. They will keep you on track!

# Practise to master

SECTION A

Write these expressions in the form $$~a(x+p)^2+q~$$.

01) $$~2x^2+4x+5~$$

Factorise (two terms): $$~2[x^2+2x]+5~$$
Complete square inside square brackets:
⇒ Half the coefficient of $$~x~$$ is $$~+1~$$
⇒ $$~(+1)^2=1~$$
⇒ So we have $$~(x+1)^2-1~$$
Substitute into square brackets: $$~2[(x+1)^2-1]+5~$$
Expand square brackets: $$~2(x+1)^2-2+5~$$
Simplify: $$~2(x+1)^2+3~$$

02) $$~2x^2+12x-3~$$

Factorise (two terms): $$~2[x^2+6x]-3~$$
Complete square inside square brackets:
⇒ Half the coefficient of $$~x~$$ is $$~+3~$$
⇒ $$~(+3)^2=9~$$
⇒ So we have $$~(x+3)^2-9~$$
Substitute into square brackets: $$~2[(x+3)^2-9]-3~$$
Expand square brackets: $$~2(x+3)^2-18-3~$$
Simplify: $$~2(x+3)^2-21~$$

03) $$~2x^2-8x+1~$$

Factorise (two terms): $$~2[x^2-4x]+1~$$
Complete square inside square brackets:
⇒ Half the coefficient of $$~x~$$ is $$~-2~$$
⇒ $$~(-2)^2=4~$$
⇒ So we have $$~(x-2)^2-4~$$
Substitute into square brackets: $$~2[(x-2)^2-4]+1~$$
Expand square brackets: $$~2(x-2)^2-8+1~$$
Simplify: $$~2(x-2)^2-7~$$

04) $$~3x^2-6x-7~$$

Factorise (two terms): $$~3[x^2-2x]-7~$$
Complete square inside square brackets:
⇒ Half the coefficient of $$~x~$$ is $$~-1~$$
⇒ $$~(-1)^2=1~$$
⇒ So we have $$~(x-1)^2-1~$$
Substitute into square brackets: $$~3[(x-1)^2-1]-7~$$
Expand square brackets: $$~3(x-1)^2-3-7~$$
Simplify: $$~3(x-1)^2-10~$$

05) $$~3x^2+18x+4~$$

Factorise (two terms): $$~3[x^2+6x]+4~$$
Complete square inside square brackets:
⇒ Half the coefficient of $$~x~$$ is $$~+3~$$
⇒ $$~(+3)^2=9~$$
⇒ So we have $$~(x+3)^2-9~$$
Substitute into square brackets: $$~3[(x+3)^2-9]+4~$$
Expand square brackets: $$~3(x+3)^2-27+4~$$
Simplify: $$~3(x+3)^2-23~$$

06) $$~3x^2+12x-2~$$

Factorise (two terms): $$~3[x^2+4x]-2~$$
Complete square inside square brackets:
⇒ Half the coefficient of $$~x~$$ is $$~+2~$$
⇒ $$~(+2)^2=4~$$
⇒ So we have $$~(x+2)^2-4~$$
Substitute into square brackets: $$~3[(x+2)^2-4]-2~$$
Expand square brackets: $$~3(x+2)^2-12-2~$$
Simplify: $$~3(x+2)^2-14~$$

07) $$~2x^2-16x+9~$$

Factorise (two terms): $$~2[x^2-8x]+9~$$
Complete square inside square brackets:
⇒ Half the coefficient of $$~x~$$ is $$~-4~$$
⇒ $$~(-4)^2=16~$$
⇒ So we have $$~(x-4)^2-16~$$
Substitute into square brackets: $$~2[(x-4)^2-16]+9~$$
Expand square brackets: $$~2(x-4)^2-32+9~$$
Simplify: $$~2(x-4)^2-23~$$

08) $$~3x^2-24x-8~$$

Factorise (two terms): $$~3[x^2-8x]-8~$$
Complete square inside square brackets:
⇒ Half the coefficient of $$~x~$$ is $$~-4~$$
⇒ $$~(-4)^2=16~$$
⇒ So we have $$~(x-4)^2-16~$$
Substitute into square brackets: $$~3[(x-4)^2-16]-8~$$
Expand square brackets: $$~3(x-4)^2-48-8~$$
Simplify: $$~3(x-4)^2-56~$$

09) $$~4x^2+8x+5~$$

Factorise (two terms): $$~4[x^2+2x]+5~$$
Complete square inside square brackets:
⇒ Half the coefficient of $$~x~$$ is $$~+1~$$
⇒ $$~(+1)^2=1~$$
⇒ So we have $$~(x+1)^2-1~$$
Substitute into square brackets: $$~4[(x+1)^2-1]+5~$$
Expand square brackets: $$~4(x+1)^2-4+5~$$
Simplify: $$~4(x+1)^2+1~$$

10) $$~5x^2+20x-6~$$

Factorise (two terms): $$~5[x^2+4x]-6~$$
Complete square inside square brackets:
⇒ Half the coefficient of $$~x~$$ is $$~+2~$$
⇒ $$~(+2)^2=4~$$
⇒ So we have $$~(x+2)^2-4~$$
Substitute into square brackets: $$~5[(x+2)^2-4]-6~$$
Expand square brackets: $$~5(x+2)^2-20-6~$$
Simplify: $$~5(x+2)^2-26~$$

SECTION B

Again, the only thing that makes this section harder is the presence of fractions. Your most valuable assets here will be concentration and patience.

Example

Express $$~2x^2+11x-5~$$ in the form $$~a(x+p)^2+q~$$

Step 1

Factorise the first two terms to a single bracket by taking out a factor of $$~a~$$. Here, $$~a~$$ (the coefficient of $$~x^2~$$) is 2 so we take out a factor of 2. Normally, we'd say these two terms can't be factorised to a single bracket because there is no common factor of 2 and 11. But when it comes to completing the square, we'll settle for a fraction!

$$~2\Big[x^2+\frac{11}{2}~x\Big]-5~$$

Step 2

Now complete the square on the bit in the square brackets.

Half the coefficient of $$~x~$$ is $$~+\frac{11}{4}~$$

$$~\Big(+\frac{11}{4}~\Big)^2=\frac{121}{16}~$$

So we have $$~\Big(x+\frac{11}{4}~\Big)^2-\frac{121}{16}$$

Step 3

Put this in the square brackets instead of $$~x^2+\frac{11}{2}~x$$.

$$~2\Big[\Big(x+\frac{11}{4}~\Big)^2-\frac{121}{16}\Big]-5~$$

Step 4

Now expand the square brackets back out by multiplying by $$~a~$$.

$$~2\Big(x+\frac{11}{4}~\Big)^2-\frac{121}{8}-5~$$

Step 5

Simplify and relax!

$$~2\Big(x+\frac{11}{4}~\Big)^2-\frac{121}{8}-\frac{40}{8}~$$

$$~2\Big(x+\frac{11}{4}~\Big)^2-\frac{161}{8}~$$

Tip!

To multiply a fraction by 2, divide the denominator by 2 (if possible); your answer will then be a fraction in its simplest form. If the denominator is not divisible by 2, you will have to multiply the numerator by 2 instead.

To divide a fraction by 2, divide the numerator by 2 (if possible); your answer will then be a fraction in its simplest form. If the numerator is not divisible by 2, you will have to multiply the denominator by 2 instead.

The same goes for multiplying and dividing a fraction by 3, 4, 5, etc.

Good luck! You'll get there!

# Practise to master

SECTION B

Write these expressions in the form $$~a(x+p)^2+q~$$.

01) $$~2x^2+6x+5$$

Factorise (two terms): $$~2[x^2+3x]+5$$
Complete square inside square brackets:
⇒ Half the coefficient of $$~x~$$ is $$~+\frac{3}{2}$$
⇒$$~\Big(+\frac{3}{2}~~\Big)^2=\frac{9}{4}$$
⇒ So we have $$~\Big(x+\frac{3}{2}~\Big)^2-\frac{9}{4}$$
Substitute into square brackets: $$~2\Big[\Big(x+\frac{3}{2}~\Big)^2-\frac{9}{4}~\Big]+5$$
Expand square brackets: $$~2\Big(x+\frac{3}{2}~\Big)^2-\frac{9}{2}~+~5$$
Simplify:
$$~2\Big(x+\frac{3}{2}~\Big)^2-\frac{9}{2}~+\frac{10}{2}$$
$$~2\Big(x+\frac{3}{2}~\Big)^2+\frac{1}{2}$$

02) $$~2x^2+2x-1$$

Factorise (two terms): $$~2[x^2+x]-1$$
Complete square inside square brackets:
⇒ Half the coefficient of $$~x~$$ is $$~+\frac{1}{2}$$
⇒$$~\Big(+\frac{1}{2}~~\Big)^2=\frac{1}{4}$$
⇒ So we have $$~\Big(x+\frac{1}{2}~\Big)^2-\frac{1}{4}$$
Substitute into square brackets: $$~2\Big[\Big(x+\frac{1}{2}~\Big)^2-\frac{1}{4}~\Big]-1$$
Expand square brackets: $$~2\Big(x+\frac{1}{2}~\Big)^2-\frac{1}{2}~-~1$$
Simplify:
$$~2\Big(x+\frac{1}{2}~\Big)^2-\frac{1}{2}~-\frac{2}{2}$$
$$~2\Big(x+\frac{1}{2}~\Big)^2-\frac{3}{2}$$

03) $$~2x^2-10x+3$$

Factorise (two terms): $$~2[x^2-5x]+3$$
Complete square inside square brackets:
⇒ Half the coefficient of $$~x~$$ is $$~-\frac{5}{2}$$
⇒$$~\Big(-\frac{5}{2}~~\Big)^2=\frac{25}{4}$$
⇒ So we have $$~\Big(x-\frac{5}{2}~\Big)^2-\frac{25}{4}$$
Substitute into square brackets: $$~2\Big[\Big(x-\frac{5}{2}~\Big)^2-\frac{25}{4}~\Big]+3$$
Expand square brackets: $$~2\Big(x-\frac{5}{2}~\Big)^2-\frac{25}{2}~+~3$$
Simplify:
$$~2\Big(x-\frac{5}{2}~\Big)^2-\frac{25}{2}~+\frac{6}{2}$$
$$~2\Big(x-\frac{5}{2}~\Big)^2-\frac{19}{2}$$

04) $$~2x^2-5x-6$$

Factorise (two terms): $$~2[x^2-\frac{5}{2}~x]-6$$
Complete square inside square brackets:
⇒ Half the coefficient of $$~x~$$ is $$~-\frac{5}{4}$$
⇒$$~\Big(-\frac{5}{4}~~\Big)^2=\frac{25}{16}$$
⇒ So we have $$~\Big(x-\frac{5}{4}~\Big)^2-\frac{25}{16}$$
Substitute into square brackets: $$~2\Big[\Big(x-\frac{5}{4}~\Big)^2-\frac{25}{16}~\Big]-6$$
Expand square brackets: $$~2\Big(x-\frac{5}{4}~\Big)^2-\frac{25}{8}~-~6$$
Simplify:
$$~2\Big(x-\frac{5}{4}~\Big)^2-\frac{25}{8}~-\frac{48}{8}$$
$$~2\Big(x-\frac{5}{4}~\Big)^2-\frac{73}{8}$$

05) $$~2x^2+9x+1$$

Factorise (two terms): $$~2[x^2+\frac{9}{2}~x]+1$$
Complete square inside square brackets:
⇒ Half the coefficient of $$~x~$$ is $$~+\frac{9}{4}$$
⇒$$~\Big(+\frac{9}{4}~~\Big)^2=\frac{81}{16}$$
⇒ So we have $$~\Big(x+\frac{9}{4}~\Big)^2-\frac{81}{16}$$
Substitute into square brackets: $$~2\Big[\Big(x+\frac{9}{4}~\Big)^2-\frac{81}{16}~\Big]+1$$
Expand square brackets: $$~2\Big(x+\frac{9}{4}~\Big)^2-\frac{81}{8}~+~1$$
Simplify:
$$~2\Big(x+\frac{9}{4}~\Big)^2-\frac{81}{8}~+\frac{8}{8}$$
$$~2\Big(x+\frac{9}{4}~\Big)^2-\frac{73}{8}$$

06) $$~2x^2+3x-2$$

Factorise (two terms): $$~2[x^2+\frac{3}{2}~x]-2$$
Complete square inside square brackets:
⇒ Half the coefficient of $$~x~$$ is $$~+\frac{3}{4}$$
⇒$$~\Big(+\frac{3}{4}~~\Big)^2=\frac{9}{16}$$
⇒ So we have $$~\Big(x+\frac{3}{4}~\Big)^2-\frac{9}{16}$$
Substitute into square brackets: $$~2\Big[\Big(x+\frac{3}{4}~\Big)^2-\frac{9}{16}~\Big]-2$$
Expand square brackets: $$~2\Big(x+\frac{3}{4}~\Big)^2-\frac{9}{8}~-~2$$
Simplify:
$$~2\Big(x+\frac{3}{4}~\Big)^2-\frac{9}{8}~-\frac{16}{8}$$
$$~2\Big(x+\frac{3}{4}~\Big)^2-\frac{25}{8}$$

07) $$~3x^2-4x+5$$

Factorise (two terms): $$~3[x^2-\frac{4}{3}~x]+5$$
Complete square inside square brackets:
⇒ Half the coefficient of $$~x~$$ is $$~-\frac{2}{3}$$
⇒$$~\Big(-\frac{2}{3}~~\Big)^2=\frac{4}{9}$$
⇒ So we have $$~\Big(x-\frac{2}{3}~\Big)^2-\frac{4}{9}$$
Substitute into square brackets: $$~3\Big[\Big(x-\frac{2}{3}~\Big)^2-\frac{4}{9}~\Big]+5$$
Expand square brackets: $$~3\Big(x-\frac{2}{3}~\Big)^2-\frac{4}{3}~+~5$$
Simplify:
$$~3\Big(x-\frac{2}{3}~\Big)^2-\frac{4}{3}~+\frac{15}{3}$$
$$~3\Big(x-\frac{2}{3}~\Big)^2+\frac{11}{3}$$

08) $$~2x^2-7x-4$$

Factorise (two terms): $$~2[x^2-\frac{7}{2}~x]-4$$
Complete square inside square brackets:
⇒ Half the coefficient of $$~x~$$ is $$~-\frac{7}{4}$$
⇒$$~\Big(-\frac{7}{4}~~\Big)^2=\frac{49}{16}$$
⇒ So we have $$~\Big(x-\frac{7}{4}~\Big)^2-\frac{49}{16}$$
Substitute into square brackets: $$~2\Big[\Big(x-\frac{7}{4}~\Big)^2-\frac{49}{16}~\Big]-4$$
Expand square brackets: $$~2\Big(x-\frac{7}{4}~\Big)^2-\frac{49}{8}~-~4$$
Simplify:
$$~2\Big(x-\frac{7}{4}~\Big)^2-\frac{49}{8}~-\frac{32}{8}$$
$$~2\Big(x-\frac{7}{4}~\Big)^2-\frac{81}{8}$$

09) $$~3x^2+8x+7$$

Factorise (two terms): $$~3[x^2+\frac{8}{3}~x]+7$$
Complete square inside square brackets:
⇒ Half the coefficient of $$~x~$$ is $$~+\frac{4}{3}$$
⇒$$~\Big(+\frac{4}{3}~~\Big)^2=\frac{16}{9}$$
⇒ So we have $$~\Big(x+\frac{4}{3}~\Big)^2-\frac{16}{9}$$
Substitute into square brackets: $$~3\Big[\Big(x+\frac{4}{3}~\Big)^2-\frac{16}{9}~\Big]+7$$
Expand square brackets: $$~3\Big(x+\frac{4}{3}~\Big)^2-\frac{16}{3}~+~7$$
Simplify:
$$~3\Big(x+\frac{4}{3}~\Big)^2-\frac{16}{3}~+\frac{21}{3}$$
$$~3\Big(x+\frac{4}{3}~\Big)^2+\frac{5}{3}$$

10) $$~4x^2+6x-5$$

Factorise (two terms): $$~4[x^2+\frac{6}{4}~x]-5$$
Complete square inside square brackets:
⇒ Half the coefficient of $$~x~$$ is $$~+\frac{3}{4}$$
⇒$$~\Big(+\frac{3}{4}~~\Big)^2=\frac{9}{16}$$
⇒ So we have $$~\Big(x+\frac{3}{4}~\Big)^2-\frac{9}{16}$$
Substitute into square brackets: $$~4\Big[\Big(x+\frac{3}{4}~\Big)^2-\frac{9}{16}~\Big]-5$$
Expand square brackets: $$~4\Big(x+\frac{3}{4}~\Big)^2-\frac{9}{4}~-~5$$
Simplify:
$$~4\Big(x+\frac{3}{4}~\Big)^2-\frac{9}{4}~-\frac{20}{4}$$
$$~4\Big(x+\frac{3}{4}~\Big)^2-\frac{29}{4}$$

11) $$~3x^2-8x+9$$

Factorise (two terms): $$~3[x^2-\frac{8}{3}~x]+9$$
Complete square inside square brackets:
⇒ Half the coefficient of $$~x~$$ is $$~-\frac{4}{3}$$
⇒$$~\Big(-\frac{4}{3}~~\Big)^2=\frac{16}{9}$$
⇒ So we have $$~\Big(x-\frac{4}{3}~\Big)^2-\frac{16}{9}$$
Substitute into square brackets: $$~3\Big[\Big(x-\frac{4}{3}~\Big)^2-\frac{16}{9}~\Big]+9$$
Expand square brackets: $$~3\Big(x-\frac{4}{3}~\Big)^2-\frac{16}{3}~+~9$$
Simplify:
$$~3\Big(x-\frac{4}{3}~\Big)^2-\frac{16}{3}~+\frac{27}{3}$$
$$~3\Big(x-\frac{4}{3}~\Big)^2+\frac{11}{3}$$

12) $$~2x^2-3x-2$$

Factorise (two terms): $$~2[x^2-\frac{3}{2}~x]-2$$
Complete square inside square brackets:
⇒ Half the coefficient of $$~x~$$ is $$~-\frac{3}{4}$$
⇒$$~\Big(-\frac{3}{4}~~\Big)^2=\frac{9}{16}$$
⇒ So we have $$~\Big(x-\frac{3}{4}~\Big)^2-\frac{9}{16}$$
Substitute into square brackets: $$~2\Big[\Big(x-\frac{3}{4}~\Big)^2-\frac{9}{16}~\Big]-2$$
Expand square brackets: $$~2\Big(x-\frac{3}{4}~\Big)^2-\frac{9}{8}~-~2$$
Simplify:
$$~2\Big(x-\frac{3}{4}~\Big)^2-\frac{9}{8}~-\frac{16}{8}$$
$$~2\Big(x-\frac{3}{4}~\Big)^2-\frac{25}{8}$$

13) $$~3x^2+5x+1$$

Factorise (two terms): $$~3[x^2+\frac{5}{3}~x]+1$$
Complete square inside square brackets:
⇒ Half the coefficient of $$~x~$$ is $$~+\frac{5}{6}$$
⇒$$~\Big(+\frac{5}{6}~~\Big)^2=\frac{25}{36}$$
⇒ So we have $$~\Big(x+\frac{5}{6}~\Big)^2-\frac{25}{36}$$
Substitute into square brackets: $$~3\Big[\Big(x+\frac{5}{6}~\Big)^2-\frac{25}{36}~\Big]+1$$
Expand square brackets: $$~3\Big(x+\frac{5}{6}~\Big)^2-\frac{25}{12}~+~1$$
Simplify:
$$~3\Big(x+\frac{5}{6}~\Big)^2-\frac{25}{12}~+\frac{12}{12}$$
$$~3\Big(x+\frac{5}{6}~\Big)^2-\frac{13}{12}$$

14) $$~4x^2+7x-6$$

Factorise (two terms): $$~4[x^2+\frac{7}{4}~x]-6$$
Complete square inside square brackets:
⇒ Half the coefficient of $$~x~$$ is $$~+\frac{7}{8}$$
⇒$$~\Big(+\frac{7}{8}~~\Big)^2=\frac{49}{64}$$
⇒ So we have $$~\Big(x+\frac{7}{8}~\Big)^2-\frac{49}{64}$$
Substitute into square brackets: $$~4\Big[\Big(x+\frac{7}{8}~\Big)^2-\frac{49}{64}~\Big]-6$$
Expand square brackets: $$~4\Big(x+\frac{7}{8}~\Big)^2-\frac{49}{16}~-~6$$
Simplify:
$$~4\Big(x+\frac{7}{8}~\Big)^2-\frac{49}{16}~-\frac{96}{16}$$
$$~4\Big(x+\frac{7}{8}~\Big)^2-\frac{145}{16}$$

15) $$~5x^2-9x+3$$

Factorise (two terms): $$~5[x^2-\frac{9}{5}~x]+3$$
Complete square inside square brackets:
⇒ Half the coefficient of $$~x~$$ is $$~-\frac{9}{10}$$
⇒$$~\Big(-\frac{9}{10}~~\Big)^2=\frac{81}{100}$$
⇒ So we have $$~\Big(x-\frac{9}{10}~\Big)^2-\frac{81}{100}$$
Substitute into square brackets: $$~5\Big[\Big(x-\frac{9}{10}~\Big)^2-\frac{81}{100}~\Big]+3$$
Expand square brackets: $$~5\Big(x-\frac{9}{10}~\Big)^2-\frac{81}{20}~+~3$$
Simplify:
$$~5\Big(x-\frac{9}{10}~\Big)^2-\frac{81}{20}~+\frac{60}{20}$$
$$~5\Big(x-\frac{9}{10}~\Big)^2-\frac{21}{20}$$