Send me best notes on surds that are uneb uce examinable

The largest perfect square factor of 72 is 36 ($36 \times 2 = 72$).

$$
\sqrt{72} = \sqrt{36 \times 2}
$$
$$
\sqrt{72} = \sqrt{36} \times \sqrt{2}
$$
$$
\sqrt{72} = \boxed{6\sqrt{2}}
$$

*   **Addition and Subtraction:** You can only add or subtract **like surds** (surds with the same number under the root sign). If surds are not alike, simplify them first to see if they become like surds.

    **Example:** Simplify $3\sqrt{5} + \sqrt{20}$.

    $$
    \sqrt{20} = \sqrt{4 \times 5} = \sqrt{4} \times \sqrt{5} = 2\sqrt{5}
    $$

    $$
    3\sqrt{5} + 2\sqrt{5} = \boxed{5\sqrt{5}}
    $$

*   **Multiplication:** Use the property $\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}$.

    **Example:** Simplify $\sqrt{6} \times \sqrt{3}$.

    $$
    \sqrt{6} \times \sqrt{3} = \sqrt{6 \times 3}
    $$
    $$
    \sqrt{6} \times \sqrt{3} = \sqrt{18}
    $$

    $$
    \sqrt{18} = \sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = \boxed{3\sqrt{2}}
    $$

*   **Division:** Use the property $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$.

    **Example:** Simplify $\frac{\sqrt{48}}{\sqrt{3}}$.

    $$
    \frac{\sqrt{48}}{\sqrt{3}} = \sqrt{\frac{48}{3}}
    $$
    $$
    \frac{\sqrt{48}}{\sqrt{3}} = \sqrt{16}
    $$

    $$
    \sqrt{16} = \boxed{4}
    $$

*   **Single Term Denominator:** If the denominator is of the form $\sqrt{a}$, multiply the numerator and denominator by $\sqrt{a}$.

    **Example:** Rationalize $\frac{5}{\sqrt{7}}$.

    $$
    \frac{5}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}}
    $$

    $$
    \frac{5\sqrt{7}}{7} = \boxed{\frac{5\sqrt{7}}{7}}
    $$

*   **Binomial Denominator:** If the denominator is of the form $a + \sqrt{b}$ or $\sqrt{a} + \sqrt{b}$, multiply the numerator and denominator by its **conjugate**. The conjugate of $a + \sqrt{b}$ is $a - \sqrt{b}$, and vice versa. This uses the difference of squares formula: $(x+y)(x-y) = x^2 - y^2$.

    **Example:** Rationalize $\frac{1}{3 - \sqrt{2}}$.

    The conjugate of $3 - \sqrt{2}$ is $3 + \sqrt{2}$.

    $$
    \frac{1}{3 - \sqrt{2}} \times \frac{3 + \sqrt{2}}{3 + \sqrt{2}}
    $$

    $$
    \frac{1 \times (3 + \sqrt{2})}{(3 - \sqrt{2})(3 + \sqrt{2})}
    $$
    $$
    \frac{3 + \sqrt{2}}{3^2 - (\sqrt{2})^2}
    $$
    $$
    \frac{3 + \sqrt{2}}{9 - 2}
    $$
    $$
    \frac{3 + \sqrt{2}}{7} = \boxed{\frac{3 + \sqrt{2}}{7}}
    $$