Simplifying surds
To simplify surds, you need to remember the square roots of perfect squares. If the number inside the root of a surd has a square number as a factor, then it can be simplified.
\(\sqrt{12} = \sqrt{4 \cdot 3} = \sqrt 4 \cdot \sqrt 3 = 2 \cdot \sqrt 3 = 2\sqrt3\)
The steps to follow to simplify surds are:
\(\sqrt{12} = \sqrt{4 \cdot 3}\)
\(\sqrt{4 \cdot 3} = \sqrt 4 \cdot \sqrt 3\)
\(\sqrt 4 \cdot \sqrt 3 = 2 \cdot \sqrt 3\)
\(2 \cdot \sqrt 3 = 2\sqrt3\)
What are the rules for using surds?
When working with surds, you need to remember the following rules:
Multiplying surds: As long as the index of the roots is the same, you can multiply surds with different numbers inside the root by simply combining them into one root and multiplying the numbers inside the root. Likewise, you can split a root into separate roots using factors.
\(\sqrt a \cdot \sqrt b = \sqrt{a \cdot b}\)
\(\sqrt2 \cdot \sqrt 5 = \sqrt{2 \cdot 5} = \sqrt{10}\)
Dividing surds: Similarly, as long as the index of the roots is the same, you can divide surds with different numbers inside the root by combining them into one root and dividing the numbers inside the root.
\(\frac{\sqrt a}{\sqrt b} = \sqrt{\frac{a}{b}} \)
\(\frac{\sqrt {10}}{\sqrt 2} = \sqrt{\frac{10}{2}} = \sqrt5\)
\(\sqrt{a} \cdot \sqrt{a} = (\sqrt{a})^2 = a\)
\(\sqrt{3} \cdot \sqrt{3} = (\sqrt{3})^2 = 3\)
\(a \cdot \sqrt{b} = \sqrt{b} \cdot a = a\sqrt{b}\)
\(3\cdot \sqrt{2} = \sqrt{2} \cdot 3 = 3\sqrt{2}\)
\(a\sqrt{x} + b\sqrt{x} = (a+b)\sqrt{x}\)
\(a\sqrt{x} - b\sqrt{x} = (a-b)\sqrt{x}\)
\(5\sqrt{3} + 2\sqrt{3} = (5+2)\sqrt{3} = 7\sqrt{3}\)
\(5\sqrt{3} - 2\sqrt{3} = (5-2)\sqrt{3} = 3\sqrt{3}\)
To add or subtract surds, you might need to simplify them first to find like terms.
You cannot add \(\sqrt{2} + \sqrt{8}\), but you can simplify \(\sqrt{8}\) first,
\(\sqrt 8 = \sqrt{4 \cdot 2} = \sqrt{4} \cdot \sqrt{2} = 2\sqrt2\) Then you can solve \(\sqrt{2} + \sqrt{8} = \sqrt{2} + 2\sqrt{2} = 3\sqrt{2}\)
\(\begin{align} 2 +\sqrt{3})(5 +\sqrt{3}) &= 2 \cdot 5 + 2\sqrt{3} + 5 \sqrt{3} + (\sqrt3)^2 \\ &=10 +7\sqrt3+3 \\ &=13 +7\sqrt3 \end{align}\)
Rationalizing the denominator of fractions containing surds
The purpose of rationalizing the denominator of fractions containing surds is to eliminate the surd from the denominator. The strategy to do this is to multiply the numerator and denominator by the surd.
Rationalize the denominator in the following expression: \(\frac{5}{\sqrt3}\)
Using the rules: \(a \cdot \sqrt{b} = \sqrt{b} \cdot a = a\sqrt b\) and \(\sqrt{a} \cdot \sqrt{a} = (\sqrt{a})^2 = a\)
\(\frac{5}{\sqrt3} = \frac{5}{\sqrt3} \cdot \frac{\sqrt3}{\sqrt3} = \frac{5\sqrt3}{(\sqrt3)^2} = \frac{5\sqrt3}{3}\)
If the denominator contains a surd and a rational number: In this case, you need to multiply the numerator and denominator by the expression in the denominator, but with the sign in the middle changed, ie if it is (+) change it to ( -) and vice versa. This expression is called the conjugate.
Rationalize the denominator in the following expression: \(\frac{(\sqrt6 + 3)}{(\sqrt6 -2)}\)
The conjugate of \((\sqrt6 - 2)\) is\((\sqrt6 + 2)\)
Multiplying the brackets and combining like terms, you can see that the surds in the denominator cancel each other.
\(\frac{(\sqrt6 + 3)}{(\sqrt6 -2)} \cdot \frac{(\sqrt6 + 2)}{(\sqrt6 +2)} = \frac{6 +2\sqrt6+3\sqrt6 +6}{6 +\cancel{2\sqrt6 -2\sqrt6}-4} = \frac{12 +5\sqrt6}{2}\)
Surds - Key takeaways
Surds are expressions that contain a square root, cube root or other roots, which produce an irrational number as a result, with infinite decimals. They are left in their root form to represent them more precisely.
To multiply and divide surds with different numbers inside the root, the index of the roots must be the same.
To add or subtract surds, the number inside the roots must be the same.
To add or subtract surds, they might need to be simplified first.
If the number inside the root of a surd has a square number as a factor, then it can be simplified.
The purpose of rationalizing the denominator of fractions containing surds is to eliminate the surd from the denominator.