Section 6.3
Radical Simplification 1
The first two rules for working with square roots are as follows.
If $a\ge 0$ and $b\ge 0$, then $\sqrt{ab}=\sqrt{a}\cdot \sqrt{b}$.
If $a\ge 0$ and $b>0$, then $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$.
Examples:
a. $\sqrt{12}=\sqrt{4\cdot 3}=\sqrt{4}\cdot \sqrt{3}=2\sqrt{3}$
b. $\sqrt{36x}=\sqrt{36}\cdot \sqrt{x}=6\sqrt{x}$
c. $\sqrt{\dfrac{18x}{25}}=\dfrac{\sqrt{9\cdot 2x}}{\sqrt{25}}=\dfrac{\sqrt{9}\cdot \sqrt{2x}}{\sqrt{25}}=\dfrac{3\sqrt{2x}}{5}$
The square root of a large number is found by factoring the large number and then using the first rule given above. Look for factors that are perfect squares.
d. $\sqrt{75600}=\sqrt{100\cdot 756}=\sqrt{100\cdot 4\cdot 189}=\sqrt{100\cdot 4\cdot 9\cdot 21}=10\cdot 2\cdot 3\sqrt{21}=60\sqrt{21}$
Standard practice is to write fractions without square roots in the denominator (it is considered preferable to have square roots in the numerator than in the denominator). One kind of simplification is illustrated below.
e. $\dfrac{2}{\sqrt{3}}=\dfrac{2}{\sqrt{3}}\cdot \dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{2\sqrt{3}}{\sqrt{9}}=\dfrac{2\sqrt{3}}{3}$
f. $\dfrac{2+4\sqrt{3}}{\sqrt{2}}=\dfrac{2+4\sqrt{3}}{\sqrt{2}}\cdot \dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{2\sqrt{2}+4\sqrt{6}}{\sqrt{4}}=\dfrac{2\left( \sqrt{2}+2\sqrt{6} \right)}{2}=\sqrt{2}+2\sqrt{6}$
In our work, we will consider all variables (i.e., any letters appearing) to represent positive numbers. Then we have the following simplifications.
g. $\sqrt{{{x}^{3}}}=\sqrt{{{x}^{2}}\cdot x}=\sqrt{{{x}^{2}}}\cdot \sqrt{x}=x\sqrt{x}$
h. $\sqrt{{{x}^{3}}{{y}^{6}}}=\sqrt{{{x}^{2}}\cdot x\cdot {{\left( {{y}^{3}} \right)}^{2}}}=\sqrt{{{x}^{2}}}\cdot \sqrt{x}\cdot \sqrt{{{\left( {{y}^{3}} \right)}^{2}}}=x{{y}^{3}}\sqrt{x}$
Simplifying radicals: Using exponent rules to simplify radicals or square roots
Exercises:
Simplify each of the following. All letters represent positive numbers.
\begin{array}{l l l}
1. \;\;\;\sqrt{50}\;\;\; & 2. \;\;\;\sqrt{75}\;\;\; & 3. \;\;\;\sqrt{200}\;\;\; \\
4. \;\;\;\sqrt{4\cdot 9\cdot 36}\;\;\; & 5. \;\;\;\sqrt{9000}\;\;\; & 6. \;\;\;\sqrt{4275}\;\;\; \\
7. \;\;\;\sqrt{401861250}\;\;\; & 8. \;\;\;\sqrt{{{2}^{2}}\cdot {{3}^{2}}\cdot {{5}^{2}}\cdot {{7}^{2}}\cdot {{9}^{2}}}\;\;\; & 9. \;\;\;\sqrt{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8\cdot 9\cdot 10}\;\;\; \\
10. \;\;\;\sqrt{\frac{4}{9}}\;\;\; & 11. \;\;\;\sqrt{\frac{8}{25}}\;\;\; & 12. \;\;\;\sqrt{\frac{64}{9}}\;\;\; \\
13. \;\;\;\sqrt{\frac{2}{3}}\;\;\; & 14. \;\;\;\sqrt{\frac{4}{3}}\;\;\; & 15. \;\;\;\sqrt{\frac{3}{5}}\;\;\; \\
16. \;\;\;\sqrt{27{{x}^{2}}}\;\;\; & 17. \;\;\;\sqrt{8{{x}^{2}}{{y}^{3}}}\;\;\; & 18. \;\;\;\sqrt{49x{{y}^{2}}{{z}^{4}}}\;\;\; \\
19. \;\;\;\sqrt{5{{x}^{5}}{{y}^{4}}}\;\;\; & 20. \;\;\;\sqrt{72{{a}^{3}}{{b}^{6}}c}\;\;\; & 21. \;\;\;\sqrt{20{{a}^{2}}{{b}^{4}}{{c}^{6}}}\;\;\; \\
22. \;\;\;\sqrt{3{{a}^{2}}b}\cdot \sqrt{27a{{b}^{3}}}\;\;\; & 23. \;\;\;\sqrt{8{{x}^{3}}y}\cdot \sqrt{32x{{y}^{3}}}\;\;\; & 24. \;\;\;\sqrt{72{{a}^{3}}{{b}^{4}}}\cdot \sqrt{2a{{b}^{4}}}\;\;\; \\
25. \;\;\;\sqrt{12{{a}^{5}}{{b}^{7}}}\cdot \sqrt{27a{{b}^{4}}}\;\;\; & 26. \;\;\;\sqrt{288{{x}^{-3}}{{y}^{-4}}}\cdot \sqrt{2x{{y}^{4}}}\;\;\; & 27. \;\;\;\sqrt{8{{x}^{5}}{{y}^{-3}}}\cdot \sqrt{2{{x}^{-1}}{{y}^{4}}}\;\;\; \\
28. \;\;\;\sqrt{12{{a}^{5}}{{b}^{7}}}\div \sqrt{27a{{b}^{4}}}\;\;\; & 29. \;\;\;\sqrt{3{{a}^{2}}b}\div \sqrt{27a{{b}^{3}}}\;\;\; & 30. \;\;\;\sqrt{8{{x}^{3}}y}\div \sqrt{32x{{y}^{3}}}\;\;\; \\
31. \;\;\;\frac{\sqrt{20{{a}^{2}}}}{2\sqrt{5a}}\;\;\; & 32. \;\;\;\frac{\sqrt{32{{a}^{3}}}}{\sqrt{8a}}\;\;\; & 33. \;\;\;\frac{\sqrt{7{{a}^{4}}}}{\sqrt{14{{a}^{5}}}}\;\;\; \\
34. \;\;\;\frac{\sqrt{10a}}{\sqrt{5{{a}^{3}}}}\;\;\; & 35. \;\;\;\frac{5}{\sqrt{5{{a}^{3}}}}\;\;\; & 36. \;\;\;\frac{\sqrt{3{{a}^{2}}b}}{\sqrt{9{{a}^{5}}{{b}^{2}}}}\;\;\; \\
37. \;\;\;\sqrt{8{{a}^{2}}{{b}^{2}}{{c}^{3}}}\;\;\; & 38. \;\;\;\sqrt{{{2}^{2}}\cdot {{3}^{3}}\cdot {{5}^{6}}\cdot 7}\;\;\; & 39. \;\;\;\frac{8\pm \sqrt{76}}{2}\;\;\; \\
40. \;\;\;\frac{6\pm \sqrt{72}}{2}\;\;\; & 41. \;\;\;\frac{4\pm \sqrt{28}}{6}\;\;\; & 42. \;\;\;\frac{-2\pm \sqrt{52}}{6}\;\;\; \\
43. \;\;\;\frac{12\pm \sqrt{140}}{2}\;\;\; & 44. \;\;\;\frac{6\pm \sqrt{8}}{14}\;\;\; & 45. \;\;\;\sqrt{24{{x}^{4}}{{y}^{7}}}\cdot \sqrt{\frac{54{{z}^{2}}y}{{{x}^{2}}}}\;\;\; \\
46. \;\;\;\sqrt{200{{x}^{6}}\cdot 800{{y}^{3}}}\;\;\; & 47. \;\;\;\sqrt{x\cdot {{x}^{2}}\cdot {{x}^{3}}\cdot {{x}^{4}}\cdot {{x}^{5}}\cdot {{x}^{6}}}\;\;\; & 48. \;\;\;\sqrt{2x}\cdot \sqrt{3{{x}^{2}}}\cdot \sqrt{4{{x}^{3}}}\cdot \sqrt{6{{x}^{4}}}\;\;\; \\
49. \;\;\;\sqrt{2\cdot {{3}^{3}}\cdot {{6}^{5}}\cdot {{8}^{7}}\cdot {{9}^{9}}\cdot {{12}^{11}}}\;\;\; & 50. \;\;\;\sqrt{x\cdot {{x}^{3}}\cdot {{x}^{5}}\cdot {{x}^{7}}\cdot {{x}^{9}}\cdot {{x}^{11}}}\;\;\; & 51. \;\;\;\sqrt{2{{x}^{2}}}\cdot \sqrt{3{{x}^{3}}}\cdot \sqrt{4{{x}^{4}}}\cdot \sqrt{5{{x}^{5}}}\;\;\;
\end{array}
52. Consider a quadratic equation in standard form $a{{x}^{2}}+bx+c=0$. Assume that $a=1$ and that $b$ is even. Since $a=1$, the Quadratic Formula gives the solutions to be $x=\frac{-b\pm \sqrt{{{b}^{2}}-4c}}{2}$. Explain why this always reduces by a factor of 2. [That's why a solution by completing the square is preferred over the Quadratic Formula in this special case.]