Solutions to Problem Set 22

Math 101-06

11-3-2017

[Radical expressions]

1. Find the exact value of $\sqrt{81}$ .

$$\sqrt{81} = 9.\quad\halmos$$


2. Find the exact value of $\sqrt{-81}$ .

$\sqrt{-81}$ is undefined.


3. Find the exact value of $-\sqrt{4}$ .

$$-\sqrt{4} = -2.\quad\halmos$$


4. Find the exact value of $\root
   3 \of 8$ .

$$\root 3 \of 8 = 2.\quad\halmos$$


5. Find the exact value of $\root
   3 \of {-8}$ .

$$\root 3 \of {-8} = -2.\quad\halmos$$


6. Find the exact value of $\sqrt{\dfrac{1}{25}}$ .

$$\sqrt{\dfrac{1}{25}} = \dfrac{1}{\sqrt{25}} = \dfrac{1}{5}.\quad\halmos$$


7. Simplify $\sqrt{54}$ .

$$\sqrt{54} = \sqrt{9 \cdot 6} = \sqrt{9} \sqrt{6} = 3 \sqrt{6}.\quad\halmos$$


8. Simplify $\sqrt{72} - 5
   \sqrt{8}$ .

$$\sqrt{72} - 5 \sqrt{8} = \sqrt{36} \sqrt{2} - 5 \sqrt{4} \sqrt{2} = 6 \sqrt{2} - 5 \cdot 2 \sqrt{2} = 6 \sqrt{2} - 10 \sqrt{2} = -4 \sqrt{2}.\quad\halmos$$


9. Multiply out and simplify: $2
   \sqrt{7} (3 + 4 \sqrt{7})$ .

$$2 \sqrt{7} (3 + 4 \sqrt{7}) = 2 \sqrt{7} \cdot 3 + 2 \sqrt{7} \cdot 4 \sqrt{7} = 6 \sqrt{7} + 8 (\sqrt{7})^2 = 6 \sqrt{7} + 8 \cdot 7 = 6 \sqrt{7} + 56.\quad\halmos$$


10. Multiply out and simplify: $(\sqrt{5} + 1)(3 \sqrt{5} - 4)$ .

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & & & $\sqrt{5}$ & & 1 & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & $3 \sqrt{5}$ & & 15 & & $3 \sqrt{5}$ & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & -4 & & $-4 \sqrt{5}$ & & -4 & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

$$(\sqrt{5} + 1)(3 \sqrt{5} - 4) = 11 - \sqrt{5}.\quad\halmos$$


11. Multiply out and simplify: $(\sqrt{11} + 3)(\sqrt{11} - 3)$ .

$$(\sqrt{11} + 3)(\sqrt{11} - 3) = (\sqrt{11})^2 - 3^2 = 11 - 9 = 2.\quad\halmos$$


12. Multiply out and simplify: $(5 - 2 \sqrt{3})(5 + 2 \sqrt{3})$ .

$$(5 - 2 \sqrt{3})(5 + 2 \sqrt{3}) = 5^2 - (2 \sqrt{3})^2 = 25 - 2^2 \cdot (\sqrt{3})^2 = 25 - 4 \cdot 3 = 25 - 12 = 13.\quad\halmos$$


13. Simplify the expression by rationalizing: $\dfrac{4 \sqrt{3}}{5 - \sqrt{3}}$ .

$$\dfrac{4 \sqrt{3}}{5 - \sqrt{3}} = \dfrac{4 \sqrt{3}}{5 - \sqrt{3}} \cdot \dfrac{5 + \sqrt{3}}{5 + \sqrt{3}} = \dfrac{4 \sqrt{3}(5 + \sqrt{3})}{(5 - \sqrt{3})(5 + \sqrt{3})} = \dfrac{20 \sqrt{3} + 4 (\sqrt{3})^2}{5^2 - (\sqrt{3})^2} =$$

$$\dfrac{20 \sqrt{3} + 4 \cdot 3}{25 - 3} = \dfrac{20 \sqrt{3} + 12}{22} = \dfrac{10 \sqrt{3} + 6}{11}.\quad\halmos$$


14. Simplify the expression by rationalizing: $\dfrac{6 + \sqrt{2}}{4 - \sqrt{2}}$ .

$$\dfrac{6 + \sqrt{2}}{4 - \sqrt{2}} = \dfrac{6 + \sqrt{2}}{4 - \sqrt{2}} \cdot \dfrac{4 + \sqrt{2}}{4 + \sqrt{2}} = \dfrac{(6 + \sqrt{2})(4 + \sqrt{2})}{(4 - \sqrt{2})(4 + \sqrt{2})} = \dfrac{26 + 10 \sqrt{2}}{4^2 - (\sqrt{2})^2} =$$

$$\dfrac{26 + 10 \sqrt{2}}{16 - 2} = \dfrac{26 + 10 \sqrt{2}}{14} = \dfrac{13 + 5 \sqrt{2}}{7}.$$

Here's the work for multiplying out the top:

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & & & 6 & & $\sqrt{2}$ & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & 4 & & 24 & & $4 \sqrt{2}$ & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & $\sqrt{2}$ & & $6 \sqrt{2}$ & & 2 & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

$$(6 + \sqrt{2})(4 + \sqrt{2}) = 26 + 10 \sqrt{2}.\quad\halmos$$


[Fractional powers]

15. Find the exact value of $25^{1/2}$ .

$$25^{1/2} = 5.\quad\halmos$$


16. Find the exact value of $8^{1/3}$ .

$$8^{1/3} = 2.\quad\halmos$$


17. Find the exact value of $(-8)^{1/3}$ .

$$(-8)^{1/3} = -2.\quad\halmos$$


18. Find the exact value of $(-16)^{1/4}$ .

An even root of a negative number is undefined:

$$(-16)^{1/4} \quad\hbox{is undefined}.\quad\halmos$$


19. Find the exact value of $-16^{1/4}$ .

In this case, you compute $16^{1/4}$ first, then negate it. So

$$16^{1/4} = 2, \quad\hbox{since}\quad 2^4 = 16.$$

Then

$$-16^{1/4} = -2.\quad\halmos$$


20. Find the exact value of $4^{5/2}$ .

$$4^{5/2} = (\sqrt{4})^5 = 2^5 = 32.\quad\halmos$$


21. Find the exact value of $27^{2/3}$ .

$$27^{2/3} = ({\root 3 \of {27}})^2 = 3^2 = 9.\quad\halmos$$


22. Find the exact value of $16^{-3/4}$ .

$$16^{-3/4} = \dfrac{1}{16^{3/4}} = \dfrac{1}{({\root 4 \of {16}})^3} = \dfrac{1}{2^3} = \dfrac{1}{8}.\quad\halmos$$


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