The square and square root of a number

In order to discuss different types of equations (see later pages) we need to know how the square and the square root of a number are defined. This should largely be a repetition.

The square of a number

The square of a number aa, written

a2a^2

(say aa squared, or aa to the power of 22) is simply defined as the multiplication of aa with itself:

a2=a⋅a\boxed{a^2= a\cdot a}

and

ca2=c⋅a⋅a\boxed{ca^2=c\cdot a\cdot a}

For example,

32=93^2=9

because 32=3⋅33^2=3\cdot 3 and

4⋅32=364\cdot 3^2=36

because 4⋅32=4⋅3⋅34\cdot 3^2=4\cdot 3\cdot 3. Since −32=−1⋅32-3^2=-1\cdot 3^2 we get that

−32=−9-3^2=-9

Also note that if we want to square a fraction, we have to use brackets:

(25)2=2525=2⋅25⋅5=425\left(\frac{2}{5}\right)^2=\frac{2}{5}\frac{2}{5}=\frac{2\cdot 2}{5\cdot 5}=\frac{4}{25}

Without bracket we read the expression as follows:

225=2⋅25=45\frac{2^2}{5}=\frac{2\cdot 2}{5}=\frac{4}{5}
Exercise 1

Calculate

  1. 3+423+4^2

  2. 3+2⋅423+2\cdot 4^2

  3. 4⋅524\cdot 5^2

  4. (32)2\left(3^2\right)^2

  5. 2(−3)22(-3)^2

  6. 2−322-3^2

  7. 562\frac{5}{6}^2

  8. (79)2\left(\frac{7}{9}\right)^2

Solution
  1. 3+42=3+16=193+4^2=3+16=19
  2. 3+2⋅42=3+2⋅16=3+32=353+2\cdot 4^2=3+2\cdot 16=3+32=35
  3. 4⋅52=4⋅25=1004\cdot 5^2=4\cdot 25=100
  4. (32)2=92=81\left(3^2\right)^2=9^2=81
  5. 2(−3)2=2⋅9=182(-3)^2=2\cdot 9=18
  6. 2−32=2−9=−72-3^2=2-9=-7
  7. 562=256\frac{5}{6}^2=\frac{25}{6}
  8. (79)2=4981\left(\frac{7}{9}\right)^2=\frac{49}{81}
Exercise 2

Resolve the brackets

  1. 2(−3)22(-3)^2

  2. (2a)2(2a)^2

  3. (x+1)2(x+1)^2

  4. (3abc)2(3abc)^2

  5. (5a2)2\left(5a^2\right)^2

Solution
  1. 2(−3)2=2⋅9=182(-3)^2=2\cdot 9=18
  2. (2a)2=2a⋅2a=2⋅2⋅a⋅a=4a2(2a)^2 =2a\cdot 2a=2\cdot 2\cdot a\cdot a=4a^2
  3. (x+1)2=(x+1)(x+1)=x2+2x+1(x+1)^2=(x+1)(x+1)=x^2+2x+1
  4. (3abc)2=3abc⋅3abc=3⋅3⋅a⋅a⋅b⋅b⋅c⋅c=9a2b2c2\left(3abc\right)^2=3abc\cdot 3abc=3\cdot 3\cdot a\cdot a\cdot b\cdot b\cdot c\cdot c = 9a^2b^2c^2
  5. (5a2)2=5a2⋅5a2=5⋅5⋅a⋅a⋅a⋅a=25a4\left(5a^2\right)^2=5a^2\cdot 5a^2=5\cdot 5\cdot a\cdot a \cdot a\cdot a= 25a^4

The square root of a number

The square root of a number aa, written

a\sqrt{a}

is defined as the positive value whose square is aa:

a=b if b2=a and b≥0\boxed{\sqrt{a}=b\, \text{if } b^2=a \text{ and } b\geq 0}

For example,

9=3\sqrt{9}=3

because 32=93^2=9. Note that if we square −3-3, we also get 99, but since we define the root of a number as a positive number, −3-3 is ignored.

Exercise 3

If possible, resolve the following square roots. Justify the result.

  1. 25\sqrt{25}

  2. 121\sqrt{121}

  3. −4\sqrt{-4}

  4. a2\sqrt{a^2}

  5. 4a2\sqrt{4a^2}

  6. 925\sqrt{\frac{9}{25}}

  7. x2+1\sqrt{x^2+1}

Solution
  1. 25=5\sqrt{25}=5 because 52=255^2=25
  2. 121=11\sqrt{121} = 11 because 112=12111^2=121
  3. −4\sqrt{-4} no solution because no number raised to the power of two is negative.
  4. a2=a\sqrt{a^2}=a because aa squared is a2a^2
  5. 4a2=2a\sqrt{4a^2}=2a because (2a)2=4a2(2a)^2=4a^2
  6. 925=35\sqrt{\frac{9}{25}}=\frac{3}{5} because 35⋅35=925\frac{3}{5}\cdot \frac{3}{5}=\frac{9}{25}
  7. x2+1\sqrt{x^2+1} not possible to resolve the root!

Square and square root cancel each other out

Taking the square of a number aa, and then taking the root of the result, gives again this number aa:

a2=a\boxed{\sqrt{a^2}=a}

We can also first the square root of aa, and if we square the result, we get again aa:

(a)2=a\boxed{\left(\sqrt{a}\right)^2=a}

So, square and square root cancel each other out.

For example, we have 42=4\sqrt{4^2}=4 and (4)2=4\left(\sqrt{4}\right)^2=4, as can be quickly verified by actually calculating the left side:

42=16=4\sqrt{4^2}=\sqrt{16}=4 (4)2=22=4\left(\sqrt{4}\right)^2=2^2=4

Click right if you want to see a general argument why they cancel each other out.

Show

a2=a\sqrt{\color{green}{a^2}}={\color{red}a} because a2=a2{\color{red}a}^2=\color{green}{a^2}

(a)2=a\left(\sqrt{a}\right)^2=a because for a number bb with a=b\sqrt{a}=b it is b2=ab^2=a.

Exercise 4

Simplify:

  1. (201)2\left(\sqrt{201}\right)^2

  2. 932\sqrt{93^2}

  3. (x2+1)2\left(\sqrt{x^2+1}\right)^2

  4. (x+1)2\sqrt{(x+1)^2}

  5. 22\sqrt{2}\sqrt{2}

  6. xxxx\sqrt{x}\sqrt{x}\sqrt{x}\sqrt{x}

Solution
  1. (201)2=201\left(\sqrt{201}\right)^2=201
  2. 932=93\sqrt{93^2}=93
  3. (x2+1)2=x2+1\left(\sqrt{x^2+1}\right)^2=x^2+1
  4. (x+1)2=x+1\sqrt{(x+1)^2}=x+1
  5. 22=(2)2=2\sqrt{2}\sqrt{2}=\left(\sqrt{2}\right)^2=2
  6. xxxx=(x)2(x)2=xx=x2\sqrt{x}\sqrt{x}\sqrt{x}\sqrt{x}=\left(\sqrt{x}\right)^2 \left(\sqrt{x}\right)^2 = x x = x^2