Question

an expression that contains a radical is in simplest radical form if (1) no number under a radical sign has a factor that is a perfect square, (2) no fraction is under a radical sign, and (3) no radical is in a denominator. use the quotient rule. then simplify the numerator and denominator. circle the result if it is in simplest radical form. $sqrt{\frac{12}{25}}=\frac{sqrt{12}}{sqrt{25}}=\frac{2sqrt{3}}{5}$ $\frac{3sqrt{2}}{10}$ $\frac{5sqrt{3}}{2}$ $\frac{5sqrt{12}}{3}$ $\frac{sqrt{3}}{5}$ is not in simplest radical form, because the number under the radical sign has 4 as a factor, and 4 is a perfect square. $\frac{2sqrt{3}}{5}$ is in simplest radical form, because the number under the radical sign, and no radical is in the denominator. circle the numbers which are in simplest radical form. $\frac{3sqrt{7}}{8}$ $sqrt{\frac{3}{4}}=$ $sqrt{\frac{3}{5}}=$ $sqrt{\frac{1}{2}}=$ $sqrt{\frac{25}{48}}=$ $sqrt{\frac{8}{9}}=$ $sqrt{\frac{4}{27}}=$ $sqrt{\frac{45}{4}}=$ $sqrt{\frac{1}{12}}=$ $sqrt{\frac{24}{49}}=$

Explanation:

Step1: Recall quotient - rule of square - roots

$\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}},a\geq0,b > 0$

Step2: Simplify numerator and denominator radicals

For example, if $a = 12$, $\sqrt{12}=\sqrt{4\times3}=2\sqrt{3}$; if $a = 27$, $\sqrt{27}=\sqrt{9\times3}=3\sqrt{3}$

Step3: Rationalize the denominator if needed

If the denominator is a non - square - number under the square - root, multiply both numerator and denominator by the square - root in the denominator. For example, for $\frac{\sqrt{3}}{\sqrt{5}}$, multiply by $\frac{\sqrt{5}}{\sqrt{5}}$ to get $\frac{\sqrt{15}}{5}$; for $\frac{1}{\sqrt{3}}$, multiply by $\frac{\sqrt{3}}{\sqrt{3}}$ to get $\frac{\sqrt{3}}{3}$

Answer:

1. $\sqrt{\frac{25}{48}}=\frac{5}{4\sqrt{3}}=\frac{5\sqrt{3}}{12}$
2. $\sqrt{\frac{1}{2}}=\frac{\sqrt{1}}{\sqrt{2}}=\frac{\sqrt{2}}{2}$
3. $\sqrt{\frac{3}{5}}=\frac{\sqrt{3}}{\sqrt{5}}=\frac{\sqrt{15}}{5}$
4. $\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$
5. $\sqrt{\frac{12}{25}}=\frac{\sqrt{12}}{\sqrt{25}}=\frac{2\sqrt{3}}{5}$
6. $\sqrt{\frac{4}{27}}=\frac{\sqrt{4}}{\sqrt{27}}=\frac{2}{3\sqrt{3}}=\frac{2\sqrt{3}}{9}$
7. $\sqrt{\frac{15}{4}}=\frac{\sqrt{15}}{2}$
8. $\sqrt{\frac{1}{12}}=\frac{\sqrt{1}}{\sqrt{12}}=\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{6}$
9. $\sqrt{\frac{8}{9}}=\frac{\sqrt{8}}{\sqrt{9}}=\frac{2\sqrt{2}}{3}$