4. Addition and Subtraction of Radicals

In algebra, we can combine terms that are similar eg.

2a + 3a = 5a

8x2 + 2x − 3x2 = 5x2 + 2x

Similarly for surds, we can combine those that are similar. They must have the same radicand (number under the radical) and the same index (the root that we are taking).

Example 1

(a) 2√7 − 5√7 + √7

Answer

In this question, the radicand (the number under the square root) is 7 in each item, and the index is 2 (that is, we are taking square root) in each item, so we can add and subtract the like terms as follows:

2√7 − 5√7 + √7 = −2√7

What I did (in my head) was to factor out √7 as follows:

2√7 − 5√7 + √7

= (2 − 5 + 1)√7

= −2√7

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(b) `root(5)6+4root(5)6-2root(5)6`

Answer

Once again, each item has the same radicand (`6`) and the same index (`5`), so we can collect like terms as follows:

`root(5)6+4root(5)6-2root(5)6=3root(5)6`

(c) `sqrt5+2sqrt3-5sqrt5`

Answer

In this example, the like terms are the √5 and −√5 (same radicand, same index), so we can add them, but the √3 term has a different radicand and so we cannot do anything with it.

`sqrt5+2sqrt3-5sqrt5=2sqrt3-4sqrt5`

Continues below

Example 2

(a) `6sqrt7-sqrt28+3sqrt63`

Answer

In each part, we are taking square root, but the number under the square root is different. We need to simplify the radicals first and see if we can combine them. We recognize that there is a √7 term involved in each item.

`6sqrt(7) - sqrt(28) + 3sqrt(63) `

`= 6sqrt(7) - sqrt(4 times 7) + 3sqrt(9 times 7)`

`= 6sqrt(7) - 2sqrt(7) + 3(3 sqrt(7))`

`= 6sqrt(7) - 2sqrt(7) + 9sqrt(7)`

`=13sqrt(7)`

(b) `3sqrt125-sqrt20+sqrt27`

Answer

We simplify the radicals first and then collect together like terms.

`3sqrt(125) - sqrt(20) + sqrt(27)`

`= 3sqrt(25 times 5) - sqrt(4 times 5) + sqrt(9 times 3)`

`=3(5sqrt(5)) - 2 sqrt(5) + 3sqrt(3)`

`=13sqrt(5) + 3sqrt(3)`

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Example 3

Simplify:

`sqrt(2/(3a))-2sqrt(3/(2a)`

Answer

Our aim here is to remove the radicals from the denominator of each fraction and then to combine the terms into one expression.

First, we multiply top and bottom of each fraction with their respective denominators. This gives us a perfect square in the denominator in each case, and we can remove the radical.

`sqrt(frac{2}{3a}) - 2 sqrt(frac{3}{2a})`

` = sqrt(frac{2(3a)}{3a(3a)}) - 2 sqrt(frac{3(2a)}{2a(2a)})`

` = sqrt(frac{6a}{9a^2}) - 2sqrt(frac{6a}{4a^2})`

We then simplify and see that we have like terms (`sqrt(6a)`).

`= frac{1}{3a}sqrt(6a) - frac{2}{2a}sqrt(6a)`

`=frac{1}{3a} sqrt(6a) - frac{1}{a} sqrt(6a)`

We then proceed to subtract the fractions by finding a common denominator (`3a`).

`= frac{sqrt(6a) - 3sqrt(6a)}{3a}`

`=frac{-2sqrt(6a)}{3a}`

`=-frac{2}{3a} sqrt(6a)`

Exercises

Q1 `sqrt7+sqrt63`

Answer

`sqrt(7) + sqrt(63)`

` = sqrt(7) + sqrt(9 times 7)`

`=sqrt(7) + 3sqrt(7)`

`=4sqrt(7)`

Q2 `2sqrt44-sqrt99+sqrt2sqrt88`

Answer

`2sqrt(44) - sqrt(99) + sqrt(2) sqrt(88)`

`=2sqrt(4 times 11) - sqrt(9 times 11) +` ` sqrt(2) sqrt(4 times 2 times 11)`

`=2(2)sqrt(11) - (3)sqrt(11) +` ` sqrt(2)(2)sqrt(2)sqrt(11)`

`=4sqrt(11) - 3sqrt(11) + 4sqrt(11)`

`=5sqrt(11)`

Q3 `root(6)sqrt2-root(12)(2^13)`

Answer

This looks ugly, but don't panic.

For the first item, finding the 6th root of a square root is the same as finding the 12th root. We need to use this rule from before:

`rootmrootn(a) = root(mn)(a)`

So the first term is:

`root(6)(sqrt(2)) = root(6 times 2)(2) = root(12)(2)`

For the second term, we need to split up the `2^13` as follows:

`2^13= 2^12× 2`

We do this so that we end up with a "12th root of 2" term so that we can simplify the final answer. We also need the following identity for this part:

`root(n)(a^n) = ( rootn(a))^n = rootn((a^n)) = a`

We'll use this in the second line, to simplify the 12th root of `2^12`:

`root(12)(2^13)`

` = root(12)(2^12 times 2)`

`=root(12)(2^12) times root(12)(2)`

`=2 root(12)(2)`

Now let's put this all together and get the final answer:

`root(6)sqrt(2) - root(12)(2^13)`

` = root(12)(2) - 2 root(12)(2)`

`= - root(12)(2)`

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