Solutions to Problem Set 23

Math 101-06

11-6-2017

[Fractional powers]

1. Simplify the expression $(x^{10/3} y^{14/5})^{1/2}$ and write your answer using positive powers. Assume the variables represent nonnegative numbers.

$$(x^{10/3} y^{14/5})^{1/2} = (x^{10/3})^{1/2} (y^{14/5})^{1/2} = x^{5/3} y^{7/5}.\quad\halmos$$


2. Simplify the expression $(-125
   x^{9/5} y^{-15})^{1/3} (x^{-2} y^{10})^{1/5}$ and write your answer using positive powers. Assume the variables represent nonnegative numbers.

$$(-125 x^{9/5} y^{-15})^{1/3} (x^{-2} y^{10})^{1/5} = (-125)^{1/3} (x^{9/5})^{1/3} (y^{-15})^{1/3} (x^{-2})^{1/5} (y^{10})^{1/5} =$$

$$-5 x^{3/5} y^{-5} x^{-2/5} y^2 = -5 x^{1/5} y^{-3} = \dfrac{-5 x^{1/5}}{y^3}.\quad\halmos$$


3. Simplify the expression $\dfrac{(6 x^{9/10})^2 y^{11/4}}{4 (x^{7/5} y^{-3/8})^2}$ and write your answer using positive powers. Assume the variables represent nonnegative numbers.

$$\dfrac{(6 x^{9/10})^2 y^{11/4}}{4 (x^{7/5} y^{-3/8})^2} = \dfrac{6^2 (x^{9/10})^2 y^{11/4}}{4 (x^{7/5})^2 (y^{-3/8})^2} = \dfrac{36 x^{9/5} y^{11/4}}{4 x^{14/5} y^{-3/4}} =$$

$$\dfrac{36 x^{9/5} y^{11/4}}{4 x^{14/5} y^{-3/4}} = 9 x^{-1} y^{7/2} = \dfrac{9 y^{7/2}}{x}.\quad\halmos$$


4. Simplify the expression $\dfrac{15 (x^{3/5} y^{3/4})^2}{(125 x^{21/5} y^{3/2})^{1/3}}$ and write your answer using positive powers. Assume the variables represent nonnegative numbers.

$$\dfrac{15 (x^{3/5} y^{3/4})^2}{(125 x^{21/5} y^{3/2})^{1/3}} = \dfrac{15 (x^{3/5})^2 (y^{3/4})^2} {125^{1/3} (x^{21/5})^{1/3} (y^{3/2})^{1/3}} = \dfrac{15 x^{6/5} y^{3/2}}{5 x^{7/5} y^{1/2}} =$$

$$3 x^{-1/5} y = \dfrac{3 y}{x^{1/5}}.\quad\halmos$$


[Radical equations]

5. Solve for x:

$$\sqrt{x + 4} = 7.$$

$$\eqalign{ \sqrt{x + 4} & = 7 \cr (\sqrt{x + 4})^2 & == 7^2 \cr x + 4 & = 7^2 \cr x & = 45 \cr}$$

Check:

$$\sqrt{45 + 4} = 7.$$

The solution is 45.


6. Solve for x:

$$\sqrt{2 x - 7} = 5.$$

$$\eqalign{ \sqrt{2 x - 7} & = 5 \cr (\sqrt{2 x - 7})^2 & == 5^2 \cr 2 x - 7 & = 5^2 \cr 2 x & = 32 \cr x & = 16 \cr}$$

Check:

$$\sqrt{2 \cdot 16 - 7} = 5.$$

The solution is 16.


7. Solve for x:

$$\sqrt{3 x + 1} = -8.$$

A radical can't be negative, so the equation has no solutions.


8. Solve the following equation for x:

$$\sqrt{2 x + 26} = x + 9.$$

$$\eqalign{ \sqrt{2 x + 26} & = x + 9 \cr (\sqrt{2 x + 26})^2 & = (x + 9)^2 \cr 2 x + 26 & = x^2 + 18 x + 81 \cr 0 & = x^2 + 16 x + 55 \cr 0 & = (x + 5)(x + 11) \cr}$$

The possible solutions are $x =
   -5$ and $x = -11$ . Check:

$$\matrix{ & \sqrt{2 x + 26} & x + 9 \cr \noalign{\vskip2pt \hrule \vskip2pt} x = -5 & 4 & 4 \cr x = -11 & 2 & -2 \cr}$$

$x = -5$ checks and $x =
   -11$ doesn't check.

The only solution is $x = -5$ .


9. Solve the following equation for x:

$$\sqrt{x + 18} = \sqrt{x + 11} + 1.$$

$$\eqalign{ \sqrt{x + 18} & = \sqrt{x + 11} + 1 \cr (\sqrt{x + 18})^2 & = (\sqrt{x + 11} + 1)^2 \cr x + 18 & = x + 11 + 2 \sqrt{x + 11} + 1 \cr x + 18 & = x + 12 + 2 \sqrt{x + 11} \cr 6 & = 2 \sqrt{x + 11} \cr 3 & = \sqrt{x + 11} \cr 3^2 & = (\sqrt{x + 11})^2 \cr 9 & = x + 11 \cr -2 & = x \cr}$$

Check the answer:

$$\sqrt{-2 + 18} = \sqrt{16} = 4.$$

$$\sqrt{-2 + 11} + 1 = \sqrt{9} + 1 = 4.$$

The solution is $x = -2$ .


10. Solve the following equation for x:

$$\sqrt{x + 7} = \sqrt{x + 6} + 1.$$

$$\eqalign{ \sqrt{x + 7} & = \sqrt{x + 6} + 1 \cr (\sqrt{x + 7})^2 & = (\sqrt{x + 6} + 1)^2 \cr x + 7 & = x + 6 + 2 \sqrt{x + 6} + 1 \cr x + 7 & = x + 7 + 2 \sqrt{x + 6} \cr 0 & = 2 \sqrt{x + 6} \cr 0 & = \sqrt{x + 6} \cr 0^2 & = (\sqrt{x + 6})^2 \cr 0 & = x + 6 \cr -6 & = x \cr}$$

Check the answer:

$$\sqrt{-6 + 7} = \sqrt{1} = 1.$$

$$\sqrt{-6 + 6} + 1 = \sqrt{0} + 1 = 1.$$

The solution is $x = -6$ .


11. Solve $\sqrt{x + 6} - 6 =
   x$ .

Start by getting the square root term by itself:

$$\eqalign{ \sqrt{x + 6} - 6 & = x \cr \sqrt{x + 6} & = x + 6 \cr}$$

Now square both sides and solve:

$$\eqalign{ (\sqrt{x + 6})^2 & = (x + 6)^2 \cr x + 6 & = x^2 + 12 x + 36 \cr 0 & = x^2 + 11 x + 30 \cr 0 & = (x + 5)(x + 6) \cr}$$

The possible solutions are $x =
   -5$ and $x = -6$ . Check them by plugging them into the left and right sides of the original equation:

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & & & $\sqrt{x + 6} - 6$ & & x & & Conclusion & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & $x = -5$ & & $\sqrt{1} - 6 = -5$ & & -5 & & Checks & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & $x = -6$ & & $\sqrt{0} - 6 = -6$ & & -6 & & Checks & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The solutions are $x = -5$ and $x = -6$ .


12. Solve $\sqrt{2 x + 2} =
   \sqrt{x + 2} + 1$ .

$$\eqalign{ \sqrt{2 x + 2} & = \sqrt{x + 2} + 1 \cr (\sqrt{2 x + 2})^2 & = (\sqrt{x + 2} + 1)^2 \cr 2 x + 2 & = (\sqrt{x + 2})^2 + 2 \sqrt{x + 2} + 1 \cr 2 x + 2 & = x + 2 + 2 \sqrt{x + 2} + 1 \cr 2 x + 2 & = x + 3 + 2 \sqrt{x + 2} \cr x - 1 & = 2 \sqrt{x + 2} \cr (x - 1)^2 & = (2 \sqrt{x + 2})^2 \cr (x - 1)^2 & = 2^2 (\sqrt{x + 2})^2 \cr x^2 - 2 x + 1 & = 4(x + 2) \cr x^2 - 2 x + 1 & = 4 x + 8 \cr x^2 - 6 x - 7 & = 0 \cr (x - 7)(x + 1) & = 0 \cr}$$

The possible solutions are $x =
   7$ and $x = -1$ .. Check them by plugging them into the left and right sides of the original equation:

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & & & $\sqrt{2 x + 2}$ & & $\sqrt{x + 2} + 1$ & & Conclusion & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & $x = 7$ & & $\sqrt{14 + 2} = 4$ & & $\sqrt{9} + 1 = 4$ & & Checks & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & $x = -1$ & & $\sqrt{-2 + 2} = 0$ & & $\sqrt{1} + 1 = 2$ & & Doesn't check & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The only solution is $x = 7$ .


He who has not lost his head over some things has no head to lose. - Jean-Paul Richter


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