A square root of a number \(a\) is a number x such that \(x^{2}=a\).
\(a\) | \(x^{2}=a\) |
\(a>0\) | \(x=\pm\sqrt{a}\) |
\(a=0\) | \(x=0\) |
\(a<0\) | None |
Calculation
Let \(a>0\) and \(b>0\).
\((\sqrt{a})^{2}=a\) |
\(\sqrt{a^{2}}=|a|\) |
\(m\sqrt{a}+n\sqrt{a}=(m+n)\sqrt{a}\) |
\(\sqrt{a}\sqrt{b}=\sqrt{ab}\) |
\(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\) |
Remark: \(\sqrt{a}\pm\sqrt{b}\not=\sqrt{a\pm b}\) |
Rationalization of the Denominator
Let \(a>0\), \(b>0\) and \(a\not= b\).
Calculation | Formula |
\(\frac{1}{\sqrt{a}}=\frac{\sqrt{a}}{\sqrt{a}\sqrt{a}}=\frac{\sqrt{a}}{a}\) | \(a\cdot a=a^{2}\) |
\(\frac{1}{\sqrt{a}\pm\sqrt{b}}=\frac{\sqrt{a}\mp\sqrt{b}}{(\sqrt{a}\pm\sqrt{b})(\sqrt{a}\mp\sqrt{b})}=\frac{\sqrt{a}\mp \sqrt{b}}{a-b}\) | \((a+b)\cdot (a-b) =a^{2}-b^{2}\) |
\(\frac{1}{\sqrt[3]{a}\pm\sqrt[3]{b}}=\frac{\sqrt[3]{a^{2}}\mp\sqrt[3]{a}\sqrt[3]{b}+\sqrt[3]{b}}{(\sqrt[3]{a}\pm\sqrt[3]{b})(\sqrt[3]{a^{2}}\mp\sqrt[3]{a}\sqrt[3]{b}+\sqrt[3]{b})}=\frac{\sqrt[3]{a^{2}}\mp\sqrt[3]{a}\sqrt[3]{b}+\sqrt[3]{b}}{a\pm b}\) | \((a+b)(a^{2}-ab+b^{2})=a^{3}+b^{3}\) \((a-b)(a^{2}+ab+b^{2})=a^{3}-b^{3}\) |
Example: Rationalize \(\frac{2}{\sqrt[3]{5}-\sqrt[3]{3}}\).
Let \(a=\sqrt[3]{5}\) and \(b=\sqrt[3]{3}\). Then
$$\frac{2}{\sqrt[3]{5}-\sqrt[3]{3}}=\frac{2(a^{2}+ab+b^{2})}{(a-b)(a^{2}+ab+b^{2})}=\frac{2(a^{2}+ab+b^{2})}{a^{3}-b^{3}}=\frac{2(\sqrt[3]{25}+\sqrt[3]{15}+\sqrt[3]{9})}{2}=\sqrt[3]{25}+\sqrt[3]{15}+\sqrt[3]{9}$$
Double Root Sign
Let \(a>0\), \(b>0\) and \(a>b\).
Calculation | Formula |
\(\sqrt{a+b\pm2\sqrt{ab}}\) \(=\sqrt{(\sqrt{a}\pm\sqrt{b})^{2}}\) \(=\sqrt{a}\pm\sqrt{b}\) | \(a^{2}+b^{2}\pm 2ab=(a\pm b)^{2}\) |
Example: \(\sqrt{8-\sqrt{60}}=\sqrt{8-2\sqrt{15}}=\sqrt{(5+3)-2\sqrt{5\times 3}}=\sqrt{(\sqrt{5}-\sqrt{3})^{2}}=\sqrt{5}-\sqrt{3}\)