Along with simplifying the form of radicals in Math, we can also carry out addition and subtraction of radicals along with multiplication too.

This page will show how to approach dealing with sum sums and situations when required.

Addition and subtraction of radicals follows the same approach of adding and subtracting like terms in Math.

If we recall:

4x + 2x = 6x     ,     4x \space {\text{--}} \space 2x + y = 2x + y.

This same principle applies to radicals.

When the ‘radicand’ and the ‘index’ are the same in two or more radicals, then they are classed as “like radicals”.

\sqrt[3]{6}   and   2{\sqrt[3]{6}}   are like radicals.

The index 3 is the same in both, as is the radicand of 6.

Examples

(1.1)

3\sqrt{2} + 4\sqrt{2} \space {\text{--}} \space 2\sqrt{2}

= \space (3 + 4 \space {\text{--}} \space 2)\sqrt{2} \space = \space 5\sqrt{2}

All radicals were like, so we were fine to add and subtract them all together.

(1.2)

\sqrt[4]{5} + 2\sqrt[4]{5} \space {\text{--}} \space 4\sqrt[4]{5}

= \space (1 + 2 \space {\text{--}} \space 4)\sqrt[4]{5} \space = \space ({\text{-}}1)\sqrt[4]{5} \space = \space {\text{-}}\sqrt[4]{5}

When the radicand and index are all the same value, we can just factor out the radical like with example (1.1) and (1.2) here, then carry out the addition or subtraction.

(1.3)

\sqrt{3} + 2\sqrt{4} + 6\sqrt{3}

= \space (1 + 6)\sqrt{3} + 2\sqrt{4} \space = \space 7\sqrt{3} + 2\sqrt{4}

The \sqrt{3} terms can be added together as the are alike, but the \sqrt{4} term can’t be added directly together with them.

(1.4)

2\sqrt{x} \space {\text{--}} \space \sqrt{y} + \sqrt{x}

= \space (2 + 1 \space)\sqrt{x} \space {\text{--}} \space \sqrt{y}

= \space 3\sqrt{x} \space {\text{--}} \space \sqrt{y}

(1.5)

5\sqrt{48} + \sqrt{92}

This addition of radicals isn’t as complicated as it may look at first.
Both radicands are multiples of 12, and we can use this to rewrite the radical terms so that a direct addition can be carried out.

5\sqrt{48} + \sqrt{92} \space\space = \space\space 5\sqrt{4 \times 12} + \sqrt{16 \times 12}

= \space\space 5\sqrt{4}\sqrt{12} + \sqrt{16}\sqrt{12}

= \space 5\times2\sqrt{12} + 4\sqrt{12} \space\space = \space\space 10\sqrt{12} + 4\sqrt{12} \space\space = \space\space 14\sqrt{12}

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