15. Multiply $(3x-7y)^2$
Solution:
Use the formula, $(a-b)^2=a^2-2ab+b^2$
$=(3x)^2-2\cdot 3x\cdot 7y+(7y)^2$
$=9x^2-42xy+49y^2$
16. Divide $\dfrac{x^2+3x-10}{x+5}$
Solution:
$=\dfrac{(x+5)(x-2)}{x+5}$
$=x-2$
17. Factor the following expressions.
(a) $64x^9y^9+24x^2y^6$
(b) $m^3+4m^2-6m-24$
(c) $8x^2-6x-9$
(d) $16x^2-81$
(e) $8c^3+125$
(f) $x^2+6x+9-4y^2$
Solution:
(a).
$=8x^2y^6(8x^7y^3+3)$
(b).
$=m^2(m+4)-6(m+4)$
$=(m+4)(m^2-6)$
18. Solve the following equations.
(a) $2k^2=9k-9$
(b) $\dfrac{3}{k+2}-\dfrac{2}{k^2-4}=\dfrac{1}{k-2}$
Solution:
(a).
Write the equation in standard form of a quadratic equation:
$2k^2-9k+9=0$
Factor:
$2k^2-6k-3k+9=0$
$2k(k-3)-3(k-3)=0$
$(k-3)(2k-3)=0$
By zero-product rule:
$k-3=0$ or $2k-3=0$
Solve for k:
$k=3$ or $k=3/2$
Solution:
$\{3, 3/2\}$
(b).
First you need to factor, whatever you can factor.
$\dfrac{3}{k+2}-\dfrac{2}{(k+2)(k-2)}=\dfrac{1}{k-2}$
Least common denominator is $(k+2)(k-2)$
Multiply this in each term:
$\dfrac{3}{k+2}\cdot (k+2)(k-2)-\dfrac{2}{(k+2)(k-2)}\cdot (k+2)(k-2)$
$=\dfrac{1}{k-2}\cdot (k+2)(k-2)$
Now, cancel out whatever you can cancel out:
$3(k-2)-2=k+3$
Now, solve for $k$:
$3k-6-2=k+3$
$2k=11$
$k=\dfrac{11}{2}$
19. Simplify the following expressions.
(a) $\dfrac{m^2-49}{m+1}\div \dfrac{7-m}{m}$
(b) $\dfrac{5x}{x^2+xy-2y^2}-\dfrac{3x}{x^2+5xy-6y^2}$
Solution:
(a).
You need to factor whatever you can factor:
$=\dfrac{(m+7)(m-7)}{m+1}\div \dfrac{7-m}{m}$
Replace the multiplication by taking the reciprocal of the rational expression that follows the division sign:
$=\dfrac{(m+7)(m-7)}{m+1}\cdot \dfrac{m}{7-m}$
$=\dfrac{(m+7)(m-7)}{m+1}\cdot \dfrac{m}{-(m-7)}$
$=-\dfrac{m(m+7)}{m+1}$
(b).
Factor the expressions in the denominators:
=$\dfrac{5x}{(x+2y)(x-y)}-\dfrac{3x}{(x+6y)(x-y)}$
LCD$=(x+2y)(x-y)(x+6y)$
Multiply and divide the LCD in each term:
=$\dfrac{5x}{(x+2y)(x-y)}\cdot \dfrac{(x+2y)(x-y)(x+6y)}{(x+2y)(x-y)(x+6y)}$
$-\dfrac{3x}{(x+6y)(x-y)}\cdot \dfrac{(x+2y)(x-y)(x+6y)}{(x+2y)(x-y)(x+6y)}$
Cancel out the common factors between the numerator and the denominator;
=$\dfrac{5x}{(x+2y)(x-y)}\cdot \dfrac{(x+2y)(x-y)(x+6y)}{(x+2y)(x-y)(x+6y)}$
$-\dfrac{3x}{(x+6y)(x-y)}\cdot \dfrac{(x+2y)(x-y)(x+6y)}{(x+2y)(x-y)(x+6y)}$
20. Simplify the following expressions.
(a) $\sqrt[5]{s^3}\:\sqrt[4]{s}$ (Write the answer in radical notation.)
(b) $\sqrt{108}$
(c) $-\sqrt[3]{-125a^6b^9c^{12}}$
(d) $3x\sqrt[3]{xy^2}-2\sqrt[3]{8x^4y^2}$
(e) $\sqrt{x^2-4x+4}$;$x\ge 0$
Solution:
(a).
First write the expression in exponential notation:
$=s^{\frac{3}{5}}\:s^{\frac{1}{4}}$
$=s^{\frac{3}{5}+\frac{1}{4}}$
$=s^{\frac{3}{5}+\frac{1}{4}}$
$=s^{\frac{17}{20}}$
Write in radical form:
$=\sqrt[20]{s^{17}}$
(b).
Factoring what is inside the radical sign:
=$\sqrt{4\cdot 9\cdot 3}$
=$2\cdot 3\sqrt{3}$
=$6\sqrt{3}$
(c).
=$-(-5)a^2b^3c^4$
=$5a^2b^3c^4$
(d).
=$3x\sqrt[3]{xy^2}-2\cdot 2 x\sqrt[3]{xy^2}$
=$3x\sqrt[3]{xy^2}-4x\sqrt[3]{xy^2}$
=$-x\sqrt[3]{xy^2}$
(e).
=$\sqrt{x^2-4x+4}$;
Factor the expression inside the radical sign:
=$\sqrt{(x-2)^2}$
=$(x-2)$
Note that, I did not put the absolute value symbol as it is given $x\ge 0$, that is $x$ is a positive number.
21. Rational the denominator for the following expressions.
(a) $\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{6}-\sqrt{5}}$
(b) $\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}$
(a).
Multiply and divide the conjugate of the denominator:
$=\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{6}-\sqrt{5}} \cdot \dfrac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}$
The denominator, you can write as a difference of squares:
$=\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{6}-\sqrt{5}} \cdot \dfrac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}$
22. Multiply.
(a) $(3\sqrt{7}+2\sqrt{5})(2\sqrt{7}-4\sqrt{5})$
(b) $(3-\sqrt{2})^2$
23. Solve the following equations.
(a) $\sqrt[3]{x-8}+3=0$
(b) $5=\sqrt{7x-3}$
(c) $(2w-1)^{2/3}-w^{1/3}=0$
24. Solve the following equations.
(a) $(t+5)^2=48$
(b) $y^2-14y+49=0$
25. Find the value of $c$ such that $9x^2-30x + c = 0$ has exactly one solution.
. For the quadratic function $f(x)=-2x^2-2x+3$, find the following:
(a) The vertex
(b) The line of symmetry
(c) The maximum or minimum value
(d) The x-intercept
(e) The y-intercept
(f) The graph of the function.
27. A club swimming pool is 30 feet long. The area of the pool is 1200 ft2. The club members want a paved walkway with uniform width around the pool. They have enough material to cover 296 ft2. How wide can the strip be?
28. Simplify as much as possible.
(a) Find $\dfrac{f}{g}$ if $f(x)=\dfrac{x^2-16}{x^2-10x+25}$ and $g(x)=\dfrac{3x-12}{x^2-3x-10}$.
(b) Find $(f-g)(x)$ if $f(x)=\dfrac{5ab}{a^2-b^2}$ and $g(x)=\dfrac{a-b}{a+b}$.
(b) Find $(f\cdot g)(-3)$ if $f(x)=\dfrac{3x}{6x^2-13x-5}$ and $g(x)=4x-10$.
29. Determine whether or not $g(x)=\sqrt{x-3}$ is one-to-one and, if possible, find $g^{-1}$.
30. Find $(f\circ g)(x)$ and $(g\circ f)(x)$ given $f(x)=4x^2-1$ and $g(x)=\dfrac{2}{x}$.
31. Simplify $\dfrac{4+3i}{5+3i}$. Write your answer in the form $a + bi$.
Solution:
Multiply and divide the complex conjugate:
$=\dfrac{(4+3i)(5-3i)}{(5+3i)(5-3i)}$
$=\dfrac{20-12i+15i-9i^2}{5^2+9^2}$
$=\dfrac{20-12i+15i-9(-1)}{25+81}$
$=\dfrac{29+3i}{106}$
$=\dfrac{29}{106}+\dfrac{3i}{106}$
32. Multiply $2i(-4-i)^2$.
Solution:
$=2i(-(4+i))^2$
$=2i(4+i)^2$
Expanding using the forumla $(a+b)^2 = a^2+2ab+b^2$:
$=2i(4^2+8i+i^2)$
$=2i(16+8i-1)$
$=2i(15+8i)$
$=30i+16i^2$
$=30i+16(-1)$
$=-16+30i$
33. Simplify.
(a) $i^{42}$
(b) $i^{17}$
(a).
Solution:
$i^{42}=i^2=-1$
Note that 2 is the remainder after dividing 42 by 4.
(b).
Solution:
$i^{17}=i^1=i$
Note that 1 is the remainder after dividing 17 by 4.
34. Solve the following equations.
(a) $16^{2x+1}=64^{x+3}$
(b) $\log_4(2x+4)=3$
(c) $2^{x+3}=5^x$
(d) $\log_2(x)+\log_2(x-7)=3$
Solution:
(a).
First, rewrite the equation so that both have same exponents:
$(4^2)^{2x+1}=(4^3)^{x+3}$
$(4)^{2(2x+1)}=(4)^{3(x+3)}$
Now, the bases are the same, so the exponents should be the same:
$2(2x+1)=3(x+3)$
Now, solve for $x$:
$4x+4=3x+9$
Solving for $x$:
$x=5$
(b).
Write the logarithmic equation in exponential form:
$4^3=2x+4$
$64=2x+4$
Solving for $x$:
$x=30$
(c).
You cannot write this equation with same base on both sides. So take log on both sides:
$\log(2^{x+3})=\log(5^x)$
$(x+3)\log(2)=x\log(5)$
$x\log(2)+3\log(2)=x\log(5)$
$x(\log(2)-\log(5))=2\log(2)$
Divide $(\log(2)-\log(5))$ to get $x$:
$x=\dfrac{2\log(2)}{\log(2)-\log(5)}$
After using the calculator:
$x=-1.513$
(d).
First combine the two logarithms on the left:
$\log_2(x(x-7))=3$
Now, write in exponential form:
$2^3=x(x-7)$
$8=x^2-7x$
This is a quadratic equation, write in standard form to solve:
$0=x^2-7x-8$
i.e., $x^2-7x-8=0$
Factoring:
$(x-7)(x+1)=0$
Applying zero-product rule:
$x-7=0$ or $x+1=0$
$x=7$ or $x=-1$
So, the solutions are
$x=-1, 7$
35. Rewrite the following expression as a single logarithm.
(e) $3\log_p(x)+\dfrac{1}{2}\log_p(y)-\dfrac{3}{2}\log_p(z)$
Solution:
Write the coefficients of log as exponents using the formula $\log_a{x^p}=p\log x$
$=\log_p(x^3)+\log_p(y^{1/2})-\log_p(z^{3/2})$
Combine them:
$=\log_p\left(\dfrac{x^3y^{1/2}}{z^{3/2}}\right)$
You can also write this as
$=\log_p\left(\dfrac{x^3\sqrt{y}}{\sqrt{z^3}}\right)$
36. Use the change of base formula to estimate the following logarithms to four decimal places.
(a) $\log_\pi(e)$
(b) $3\log_6 2.75$
Solution:
(a).
Use the change of base formula. Change the base to $e$ as there is an $e$ in the log. (You can also change the base to 10).
$\log_\pi(e)=\dfrac{\ln(e)}{\ln (\pi)}$
$=\dfrac{1}{\ln (\pi)}$
$=\dfrac{1}{1.1447}$
$=0.874$, after rounding to the thousandth.
37. Graph.
(e) $y=\left(\dfrac{2}{3}\right )^x$
(e) $y=\log_2 x$
38. Solve the following problems using the interest formulas.
(a) What will be the amount A in an account with an initial principal of $4000 if interest is compounded continuously at a rate of 3.5% for 6 years? Also, how long does it take for the account to double?
(a) A college loan of \$29,000 is made at 3% interest compounded annually. After t years, the amount due, A, is given by the function $A(t)=29,000(1.03)^t$. If no payments are made, how long will it take for the amount due to reach \$35,000?
Solution:
(a)
For the interest compounded continuously, you should use the following formula:
$A=P e^{rt}$
$t$ is the number of years, $A$ is the amount after $t$ years, $P$ is the Principal (initial deposit or loan if money is borrowed) and $r$ is the interest rate.
You are given:
$P=4000\:\$$; $r=3.5\%=3.5/100=0.035$; and $t=6$ years. You need to find the amount $A$ after $6$ years. Substituting, $P$,$r$ and $t$ in the equation:
$A=4000 e^{0.035\cdot 6}$
$A=4000 e^{0.035\cdot 6}$
$ =4000 e^{0.21}$
$ =4000 \cdot 1.2337$
$ =4934.71 \:\$$
This is one of the answer.
Next, you need to find how long, $(t)$ does it take to double the amount.
Amount is doubled, so, $A=2P$. Substituting this, $A$ and $r$ in the formula:
$2P=Pe^{0.035t}$
Divide out $P$, and solve for $t$:
$2=e^{0.035t}$
To solve, take $\ln$ on both sides:
$ln (2)= ln(e^{0.035t})$
$ln (2)= 0.035t$
$t=ln(2)/0.035=19.8$ years
(b).
In this problem, the interest is compounded annually, that is periodically. So the formula to be used is
$A=P\left(1+\dfrac{r}{n}\right)^{nt}$
where $n$ is the number of times per year, the interest is compounded.
Since the interest is compunded annually, $n=1$.
Interest rate is, $r=3\%=3/100=0.03$
You need to find the time $t$ to reach the amount, $35,000$.
Substitute, $A=35,000$ and solve for $t$:
$35000=29000(1.03)^t$
To solve this exponential equation, take log on both sides:
$\log(35000)=\log(29000(1.03)t)$
$\log(35000)=1.03t\log(29000)$
$t=\dfrac{\log(35000)}{1.03\log(29000)}=0.989$ years