Solve a quadratic inequality by first putting it in the standard form $\displaystyle ax^{2} + bx + c \ge 0$ (or similar); this is done by subtracting $\displaystyle 12x$ from both sides:

$\displaystyle x^{2} + 32 - 12x \ge 12x - 12x$

$\displaystyle x^{2} - 12x + 32 \ge 0$

Factor the quadratic trinomial on the left:

$\displaystyle x^{2} - 12x + 32 = (x + \square )(x + \square )$

We are looking for two integers whose product is 32 and whose sum is $\displaystyle -12$. Through a little trial and error, we find $\displaystyle -8$ and $\displaystyle -4$, so

$\displaystyle x^{2} - 12x + 32 = (x -8 )(x -4 )$,

making the inequality 

$\displaystyle (x -8 )(x -4 ) \ge 0$

The boundary points of the solution set are the points at which either factor is equal to 0:

$\displaystyle x- 8 = 0$, in which case $\displaystyle x= 8$, and

$\displaystyle x- 4 = 0$, in which case $\displaystyle x= 4$.

These points divide the real numbers into three intervals: $\displaystyle (-\infty, 4), (4, 8), (8, \infty)$

We can choose one test point from each interval to determine the truth of each statement on the interval:

$\displaystyle (-\infty, 4)$: choose $\displaystyle x= 0$ 

$\displaystyle x^{2} + 32 \ge 12x$

$\displaystyle 0^{2} + 32 \ge 12 \cdot 0$

$\displaystyle 0 + 32 \ge 0$

$\displaystyle 32 \ge 0$

True: include $\displaystyle (-\infty, 4)$

$\displaystyle (4, 8)$: choose $\displaystyle x= 5$

$\displaystyle x^{2} + 32 \ge 12x$

$\displaystyle 5^{2} + 32 \ge 12 \cdot 5$

$\displaystyle 25 + 32 \ge 60$

$\displaystyle 57 \ge 60$

False: exclude $\displaystyle (4, 8)$.

$\displaystyle (8, \infty)$: choose $\displaystyle x= 9$

$\displaystyle x^{2} + 32 \ge 12x$

$\displaystyle 9^{2} + 32 \ge 12 \cdot 9$

$\displaystyle 81 + 32 \ge 108$

$\displaystyle 113 \ge 108$

True: include $\displaystyle (8, \infty)$

The boundaries are also included, since the inequality allows for inclusion ("greater than or equal to"). The solution set is $\displaystyle (-\infty, 4] \cup [8, \infty)$.

The boundary points of a rational inequality are those values of the variable for which the numerator or the denominator are equal to 0:

If numerator

$\displaystyle x - 4 = 0$, then

$\displaystyle x = 4$.

Since equality is permitted in the statement ( $\displaystyle \le$ ), include this value of $\displaystyle x$.

If denominator 

$\displaystyle x + 8 = 0$,

then 

$\displaystyle x = -8$

Since this value makes the denominator 0, exclude it, regardless of the inequality symbol.

There are three intervals which must be tested for inclusion or exclusion:

$\displaystyle (\infty , -8)$, $\displaystyle (-8, 4 )$, $\displaystyle (4, \infty)$

Test one value on each interval in the original inequality for truth or falsity in order to determine which intervals should be included:

$\displaystyle (\infty , -8)$: Set $\displaystyle x = -9$

$\displaystyle \frac{x- 4}{x+ 8} \le 0$

$\displaystyle \frac{ -9 - 4}{-9 + 8} \le 0$

$\displaystyle \frac{ -13}{-1} \le 0$

$\displaystyle 13 \le 0$

False; exclude $\displaystyle (\infty , -8)$.

$\displaystyle (-8, 4 )$: Set $\displaystyle x = 0$

$\displaystyle \frac{x- 4}{x+ 8} \le 0$

$\displaystyle \frac{0- 4}{0+ 8} \le 0$

$\displaystyle \frac{-4}{8} \le 0$

$\displaystyle -\frac{1}{2} \le 0$

True; include $\displaystyle (-8, 4 )$

$\displaystyle (4, \infty)$: Set $\displaystyle x = 5$

$\displaystyle \frac{x- 4}{x+ 8} \le 0$

$\displaystyle \frac{5- 4}{5+ 8} \le 0$

$\displaystyle \frac{1}{13} \le 0$

False; exclude $\displaystyle (4, \infty)$

Since $\displaystyle -8$ is excluded as a solution and 4 is included, the correct solution set is the interval $\displaystyle (-8, 4 ]$.

Solve a quadratic inequality by first putting it in the standard form $\displaystyle ax^{2} + bx + c \ge 0$ (or similar); this is done by first applying the FOIL method to the product of the binomials on the right:

$\displaystyle (x-7)(2x+ 5)\ge 21$

$\displaystyle x \cdot 2x + x \cdot 5 -7 \cdot 2x -7 \cdot 5 \ge 21$

$\displaystyle 2x ^{2} + 5 x -14 x -35 \ge 21$

$\displaystyle 2x ^{2} -9 x -35 \ge 21$

Subtract 21 from both sides:

$\displaystyle 2x ^{2} -9 x -35 - 21 \ge 21 - 21$

$\displaystyle 2x ^{2} -9 x -56 \ge 0$

The $\displaystyle ac$ method can be used to factor the trinomial; we are looking for two integers whose product is $\displaystyle 2(-56) = -112$ and whose sum is $\displaystyle -9$; by trial and error, we find that the numbers are 7 and $\displaystyle -16$. The trinomial can be rewritten:

$\displaystyle 2x ^{2} -16x +7x -56 \ge 0$

Factoring by grouping:

$\displaystyle (2x ^{2} -16x) +(7x -56) \ge 0$

$\displaystyle 2x (x-8) + 7 (x-8) \ge 0$

$\displaystyle (2x + 7 )(x-8) \ge 0$

The boundary points of the solution set are the points at which either factor is equal to 0:

$\displaystyle 2x+7 = 0$

$\displaystyle 2x = -7$

$\displaystyle x = - \frac{7}{2} = -3 \frac{1}{2}$

and 

$\displaystyle x - 8 = 0$

$\displaystyle x = 8$

These points divide the real numbers into three intervals: $\displaystyle \left (-\infty, -3 \frac{1}{2} \right ), \left (-3 \frac{1}{2}, 8 \right ) , (8, \infty)$

Test one value on each interval in the original inequality for truth or falsity in order to determine which intervals should be included:

$\displaystyle \left (-\infty, -3 \frac{1}{2} \right )$: Set $\displaystyle x = -4$

$\displaystyle (x-7)(2x+ 5)\ge 21$

$\displaystyle (-4 -7) [2 \cdot (-4)+ 5] \ge 21$

$\displaystyle -11 (-8+ 5) \ge 21$

$\displaystyle -11 (-3) \ge 21$

$\displaystyle 33 \ge 21$

True: include $\displaystyle \left (-\infty, -3 \frac{1}{2} \right )$.

$\displaystyle \left (-3 \frac{1}{2}, 8 \right )$: Set $\displaystyle x= 0$

$\displaystyle (x-7)(2x+ 5)\ge 21$

$\displaystyle (0-7)(2 \cdot 0 + 5)\ge 21$

$\displaystyle -7( 0 + 5)\ge 21$

$\displaystyle -7( 5)\ge 21$

$\displaystyle -35 \ge 21$

False: exclude $\displaystyle \left (-3 \frac{1}{2}, 8 \right )$

$\displaystyle (8, \infty)$: Set $\displaystyle x = 9$

$\displaystyle (x-7)(2x+ 5)\ge 21$

$\displaystyle (9-7)(2 \cdot 9 + 5)\ge 21$

$\displaystyle 2(18 + 5)\ge 21$

$\displaystyle 2(23)\ge 21$

$\displaystyle 46\ge 21$

True: include $\displaystyle (8, \infty)$.

The boundaries are also included, since the inequality allows for inclusion ("greater than or equal to"). The solution set is $\displaystyle \left (-\infty, -3 \frac{1}{2} \right ] \cup [8, \infty)$.

Get all expressions in the rational inequality over to the same side by adding 4 to both sides:

$\displaystyle \frac{x-7}{x+ 3} \ge -4$

$\displaystyle \frac{x-7}{x+ 3} + 4 \ge -4 + 4$

$\displaystyle \frac{x-7}{x+ 3} + 4 \ge 0$

Rewrite the expression on the left as a single rational expression, as follows:

$\displaystyle \frac{x-7}{x+ 3} + \frac{4(x+3)}{x+3} \ge 0$

$\displaystyle \frac{x-7}{x+ 3} + \frac{4 x+4 \cdot 3 }{x+3} \ge 0$

$\displaystyle \frac{x-7}{x+ 3} + \frac{4 x+12 }{x+3} \ge 0$

$\displaystyle \frac{x-7 + 4x + 12}{x+ 3} \ge 0$

$\displaystyle \frac{x + 4x -7 + 12}{x+ 3} \ge 0$

$\displaystyle \frac{5x+5}{x+ 3} \ge 0$

The boundary points of a rational inequality are those values of the variable for which the numerator or the denominator are equal to 0:

If numerator

$\displaystyle 5x+5 = 0$

then

$\displaystyle 5x = -5$

$\displaystyle x = -1$

Since equality is permitted in the statement ( $\displaystyle \le$ ), include this value of $\displaystyle x$.

If denominator 

$\displaystyle x+ 3 = 0$

then

$\displaystyle x = -3$.

Since this value makes the denominator 0, exclude it, regardless of the inequality symbol.

There are three intervals which must be tested for inclusion or exclusion:

$\displaystyle (\infty , -3)$, $\displaystyle (-3, -1 )$, $\displaystyle (-1, \infty)$

Test one value on each interval in the original inequality for truth or falsity in order to determine which intervals should be included:

$\displaystyle (\infty , -3)$: Set $\displaystyle x = -4$

$\displaystyle \frac{x-7}{x+ 3} \ge -4$

$\displaystyle \frac{-4-7}{-4+ 3} \ge -4$

$\displaystyle \frac{-11}{-1} \ge -4$

$\displaystyle 11 \ge -4$

True; include $\displaystyle (\infty , -3)$

$\displaystyle (-3, -1 )$: Set $\displaystyle x = -2$

$\displaystyle \frac{x-7}{x+ 3} \ge -4$

$\displaystyle \frac{-2-7}{-2+ 3} \ge -4$

$\displaystyle \frac{-9}{1} \ge -4$

$\displaystyle -9 \ge 4$

False: exclude $\displaystyle (-3, -1 )$

$\displaystyle (-1, \infty)$: Set $\displaystyle x = 0$

$\displaystyle \frac{x-7}{x+ 3} \ge -4$

$\displaystyle \frac{0-7}{0+ 3} \ge -4$

$\displaystyle \frac{ -7}{ 3} \ge -4$

$\displaystyle -2 \frac{ 1}{ 3} \ge -4$

True: include $\displaystyle (-1, \infty)$.

Since $\displaystyle -3$ is excluded as a solution and $\displaystyle -1$ is included, the correct solution set is $\displaystyle (\infty , -3) \cup [-1, \infty)$

In an absolute value inequality, the absolute value expression must be isolated on one side first. We can do this by subtracting 21 from both sides:

$\displaystyle |-6x+14| + 21 < 78$

$\displaystyle |-6x+14| + 21-21 < 78 -21$

$\displaystyle |-6x+14| < 57$

This can be rewritten as the three-part inequality

$\displaystyle -57 < -6x+14 < 57$

Subtract 14 from all three expressions:

$\displaystyle -57 - 14 < -6x+14 - 14 < 57 - 14$

$\displaystyle -71 < -6x < 43$

Divide all three expressions by $\displaystyle -6$, reversing the inequality symbols since you are dividing by a negative number:

$\displaystyle -71\div (-6) >-6x \div (-6)> 43 \div (-6)$

$\displaystyle 11 \frac{5}{6} >x > -7 \frac{1}{6}$

In interval notation, this is $\displaystyle \left ( -7 \frac{1}{6}, 11 \frac{5}{6} \right )$.

$\displaystyle 42 - |-2x+19| < 14$

In an absolute value inequality, the absolute value expression must be isolated on one side first. We can di this by first subtracting 42 from both sides:

$\displaystyle 42 - |-2x+19| - 42< 14 - 42$

$\displaystyle - |-2x+19| < -28$

Divide by $\displaystyle -1$, reversing the direction of the inequality symbol since we are dividing by a negative number:

$\displaystyle \frac{- |-2x+19| }{-1}> \frac{-28}{-1}$

$\displaystyle |-2x+19| >28$

This inequality can be rewritten as the compound inequality

$\displaystyle -2x+19 < -28$ or $\displaystyle -2x+19 >28$

Solve each simple inequality separately. 

$\displaystyle -2x+19 < -28$

Subtract 19 from both sides:

$\displaystyle -2x+19 - 19 < -28 - 19$

$\displaystyle -2x < -47$

Divide by $\displaystyle -2$, remembering to reverse the symbol:

$\displaystyle \frac{-2x}{-2} > \frac{-47}{-2}$

$\displaystyle x >23.5$

In interval notation, this is $\displaystyle (23.5, \infty)$.

Carry out the same steps on the other simple inequality:

$\displaystyle -2x+19 >28$

$\displaystyle -2x+19 -19 >28 - 19$

$\displaystyle -2x >9$

$\displaystyle \frac{-2x}{-2} < \frac{9}{-2}$

$\displaystyle x < -4.5$

In interval notation, this is $\displaystyle (-\infty, -4.5)$.

Since the two simple inequalities are connected by an "or", their individual solution sets are connected by a union; the solution set is

$\displaystyle (-\infty, -4.5) \cup (23.5, \infty)$.

In general, an equation of depreciation looks like the following.

$\displaystyle y=a(1-r)^t$, where $\displaystyle a$ is the starting amount, $\displaystyle r$ is the common ratio, and $\displaystyle t$ is time.

For us $\displaystyle a=1200$, $\displaystyle r=\frac{1}{3}$, and $\displaystyle t=t$.

$\displaystyle y=1200\left(1-\frac{1}{3}\right)^t$

$\displaystyle y=1200\left(\frac{2}{3}\right)^t$

To answer this question, we should create a table of values and compare. For calculating the difference between the y value and the y value of the line of best, take the absolute value since it won't make since if it is negative.

$\displaystyle \begin{tabular}{cccc} x& y & line of best fit & difference between y and line of best fit\\ \hline 1 & 1 & 2.2053& 2.2053\\ 1 & 2& 2.2053 & 0.2053\\ 1 & 3 & 2.2053 & 0.7947\\ 2& 1 & 2.5074 & 1.5074\\ 4 & 3 & 3.1116 & 0.1116\\ 4 & 5 & 3.1116 & 1.8884\\ 4 & 6 & 3.1116 & 2.8884\\ 6 & 2 & 3.7158 & 1.7158\\ 6 & 3 & 3.7158 & 0.7158 \end{tabular}$

From our table of values, we can see that the point furthest from the line of best fit is $\displaystyle (4,6)$.

First step is to set it equal to zero

$\displaystyle f(x)=x^3-3x$

$\displaystyle x^3-3x=0$

Now factor out an $\displaystyle x$

$\displaystyle x(x^2-3)=0$

Set up the equations, and solve for $\displaystyle x$.

$\displaystyle x=0$

$\displaystyle x^2-3=0$

$\displaystyle x^2=3$

$\displaystyle x=\pm\sqrt3$