JEE Mathematics Surds and Indices Notes Set 01

Surds and Rationalisation

1. A surd is an irrational root of a rational number, a root whose exact value cannot be written as a fraction, for example \(\sqrt2,\ \sqrt3,\ \sqrt5\).
2. Laws for non-negative a and positive b: \(\sqrt{a}\,\sqrt{b}=\sqrt{ab}\), \(\dfrac{\sqrt a}{\sqrt b}=\sqrt{\dfrac ab}\), \(\sqrt{a^2b}=a\sqrt b\), and \((\sqrt a+\sqrt b)(\sqrt a-\sqrt b)=a-b\).
3. Warning: \(\sqrt a+\sqrt b\ne\sqrt{a+b}\) in general, and \(\sqrt a-\sqrt b\ne\sqrt{a-b}\).
4. Like surds (same surd part) can be added or subtracted, for example \(2\sqrt6+3\sqrt6=5\sqrt6\). To bring a surd to its simplest form, pull out the largest perfect-square factor: \(\sqrt{54}=\sqrt{9\cdot6}=3\sqrt6\).

Rationalising the Denominator

To remove a surd from a denominator, multiply the numerator and denominator by a suitable factor so that the denominator becomes rational.

Worked Examples

1. \(\dfrac1{\sqrt5}=\dfrac{\sqrt5}{5}=\dfrac{2.23606}{5}\approx0.4472\).
2. \(\dfrac{\sqrt7}{\sqrt7-\sqrt5}=\dfrac{\sqrt7(\sqrt7+\sqrt5)}{7-5}=\dfrac{7+\sqrt{35}}{2}\approx\dfrac{7+5.916}{2}=6.458\).
3. \(\dfrac{\sqrt3}{\sqrt3-\sqrt2}=\dfrac{\sqrt3(\sqrt3+\sqrt2)}{3-2}=3+\sqrt6\).
4. \(\dfrac{2\sqrt3-3\sqrt2}{2\sqrt6+5}\): multiply by \(2\sqrt6-5\). Numerator \(=12\sqrt2-10\sqrt3-12\sqrt3+15\sqrt2=27\sqrt2-22\sqrt3\); denominator \(=24-25=-1\). So the value is \(22\sqrt3-27\sqrt2\).
5. Simplify \(\sqrt{24}+\sqrt{54}-2\sqrt{96}+3\sqrt6=2\sqrt6+3\sqrt6-8\sqrt6+3\sqrt6=0\).
6. Simplify \(2\sqrt{75}-\sqrt{147}+\sqrt{50}+2\sqrt{12}-\sqrt{18}=10\sqrt3-7\sqrt3+5\sqrt2+4\sqrt3-3\sqrt2=7\sqrt3+2\sqrt2\).
7. Compare \(4\sqrt7\) and \(6\sqrt3\): bring the numbers inside the root, \(\sqrt{112}\) and \(\sqrt{108}\), so \(4\sqrt7>6\sqrt3\). Similarly \(2\sqrt3=\sqrt{12}<\sqrt{15}\).
8. \((\sqrt a-2\sqrt b)\sqrt{ab}=a\sqrt b-2b\sqrt a\).
9. \(\sqrt{x^4+x^2y^2}+\sqrt{x^2y^2+y^4}=x\sqrt{x^2+y^2}+y\sqrt{x^2+y^2}=(x+y)\sqrt{x^2+y^2}\).
10. \(\dfrac{\sqrt{x+y}+\sqrt{x-y}}{\sqrt{x+y}-\sqrt{x-y}}\): multiply by the conjugate to get \(\dfrac{(x+y)+(x-y)+2\sqrt{x^2-y^2}}{2y}=\dfrac{x+\sqrt{x^2-y^2}}{y}\).

Indeterminate Forms

\(\dfrac0c=0\) for \(c\ne0\), and \(\dfrac c0\) is undefined, but \(\dfrac00\) cannot be assigned any single value, so it is called indeterminate. For example \(\dfrac{x-5}{x-5}\) takes the form \(\dfrac00\) at \(x=5\), although it equals 1 for every \(x\ne5\).