Question 1

Solve the inequality.
-$$\frac { 14 } { x - 6 } > \frac { 12 } { x - 1 }$$
A) $( - 29,1 )$ or $( 1,6 )$
B) $( - 29,1 )$ or $( 6 , \infty )$
C) $( - \infty , - 29 )$ or $( 1,6 )$
D) $( - \infty , - 29 )$ or $( 6 , \infty )$

Accepted Answer

To solve the inequality, find the critical points by setting the numerators equal: $14(x-1) > 12(x-6)$. Solving gives $x > -29$. The denominators $x-6$ and $x-1$ cannot be zero, so $x \neq 6$ and $x \neq 1$. Test intervals around these points to find where the inequality holds: $(-29, 1)$ and $(6, \infty)$.

Question 2

Solve the inequality.
-$$\frac { ( x + 10 ) ( x - 2 ) } { x - 1 } \geq 0$$
A) $( - \infty , - 10 ]$ or $[ 2 , \infty )$
B) $[ - 10,1 )$ or $[ 2 , \infty )$
C) $( - \infty , - 10 ]$ or $( 1,2 ]$
D) $[ - 10,1 )$ or $( 1,2 ]$

Accepted Answer

The answer of Solve the inequality.
-\[\frac { ( x +...

Question 3

Solve the inequality.
-$$\frac { 15 x } { 5 - x } \geq 5 x$$
A) $[ 5 , \infty )$
B) $( - \infty , 0 ]$ or $[ 2,5 )$
C) $( - \infty , 2 ]$ or $[ 5 , \infty )$
D) $[ 0,2 ]$ or $[ 5 , \infty )$

Accepted Answer

First, simplify the inequality: $\frac{15x}{5-x} \geq 5x$ becomes $\frac{15x}{5-x} - 5x \geq 0$, which simplifies to $\frac{15x - 5x(5-x)}{5-x} \geq 0$, leading to $\frac{15x - 25x + 5x^2}{5-x} \geq 0$, and further to $\frac{5x^2 - 10x}{5-x} \geq 0$. Factoring out gives $\frac{5x(x - 2)}{5-x} \geq 0$. The critical points from the numerator are $x = 0$ and $x = 2$, and from the denominator $x = 5$. Testing intervals around these points gives the solution set as $x \in (-\infty, 0]$ or $x \in [2, 5)$.

Question 4

Solve the inequality.
-$$\frac { ( x - 2 ) ^ { 2 } } { x ^ { 2 } - 9 } > 0$$
A) $( - \infty , - 3 )$ or $( 3 , \infty )$
B) $( - \infty , - 3 )$ or $( 2,3 )$
C) $( - 3,2 )$ or $( 3 , \infty )$
D) $( - 3,2 )$ or $( 2,3 )$

Accepted Answer

The inequality is positive when the numerator and denominator have the same sign. The denominator $x^2 - 9$ is positive outside the roots $-3$ and $3$, and the numerator $(x - 2)^2$ is always non-negative. The inequality is not defined at $x = -3$ and $x = 3$, so the solution is $(-\infty, -3)$ and $(3, \infty)$.

Question 5

Solve the inequality.
-$$x + \frac { 30 } { x } < 13$$
A) $( - \infty , 0 )$ or $( 10 , \infty )$
B) $( - \infty , 0 )$ or $( 3,10 )$
C) $( 0,3 )$ or $( 3,10 )$
D) $( 0,3 )$ or $( 10 , \infty )$

Accepted Answer

The answer of Solve the inequality.
-\[x + \frac { 30...

Question 6

Solve the inequality.
-$$\frac { ( x - 3 ) ( x + 3 ) } { x } \leq 0$$
A) $[ - 3,0 )$ or $( 0,3 ]$
B) $[ - 3,0 )$ or $[ 3 , \infty )$
C) $( - \infty , - 3 ]$ or $( 0,3 ]$
D) $( - \infty , - 3 ]$ or $[ 3 , \infty )$

Accepted Answer

The answer of Solve the inequality.
-\[\frac { ( x -...

Question 7

Solve the inequality.
-$$\frac { ( x - 1 ) ( 3 - x ) } { ( x - 2 ) ^ { 2 } } \leq 0$$
A) $( - \infty , - 3 ]$ or $( - 2 , - 1 )$ or $[ 1 , \infty )$
B) $( - \infty , - 3 )$ or $( - 1 , \infty )$
C) $( - \infty , 1 )$ or $( 3 , \infty )$
D) $( - \infty , 1 ]$ or $[ 3 , \infty )$

Accepted Answer

The inequality $\frac { ( x - 1 ) ( 3 - x ) } { ( x - 2 ) ^ { 2 } } \leq 0$ can be solved by finding the zeros of the numerator and denominator and analyzing the sign changes. The zeros of the numerator are $x = 1$ and $x = 3$, and the denominator $x = 2$ does not change the sign since it's squared and always positive except at $x = 2$ where the expression is undefined. The inequality is non-positive (negative or zero) when $x$ is between the roots of the numerator, including the roots themselves since the inequality allows for equality ($\leq 0$). Therefore, the solution includes $x$ values from $-\infty$ up to $1$ (inclusive) and from $3$ to $\infty$ (inclusive), excluding $x = 2$ where the expression is undefined. This corresponds to the intervals $( - \infty , 1 ]$ and $[ 3 , \infty )$.

Question 8

Solve the inequality.
-$$\frac { x ^ { 2 } ( x - 11 ) ( x + 2 ) } { ( x - 4 ) ( x + 8 ) } \geq 0$$
A) $( - 8 , - 2 ]$ or $( 4,11 ]$
B) $( - \infty , - 8 )$ or $[ 11 , \infty )$
C) $( - \infty , - 8 )$ or $[ - 2,0 )$ or $( 0,4 )$ or $[ 11 , \infty )$
D) $( - \infty , - 8 )$ or $[ - 2,4 )$ or $[ 11 , \infty )$

Accepted Answer

The inequality is non-negative when the product of the factors is non-negative. The critical points are $x = -8, -2, 0, 4, 11$. Testing intervals between these points, the inequality holds for $( - \infty , - 8 )$, $[ - 2,4 )$, and $[ 11 , \infty )$.

Question 9

Solve the inequality.
-$$\frac { 5 x ^ { 2 } - 4 x - 9 } { x + 5 } \leq 0$$
A) $( - \infty , - 1 ]$ or $\left( \frac { 9 } { 5 } , \infty \right)$
B) $( - \infty , - 5 )$ or $\left[ - 1 , \frac { 9 } { 5 } \right]$
C) $( - 5 , - 1 ]$ or $\left( \frac { 9 } { 5 } , \infty \right)$
D) $( - \infty , - 5 ]$ or $\left[ - 1 , \frac { 9 } { 5 } \right]$

Accepted Answer

The answer of Solve the inequality.
-\[\frac { 5 x ^...

Question 10

Solve the inequality.
-$$\frac { 3 x } { 7 - x } < x$$
A) $( 4,7 )$
B) $( 7 , \infty )$
C) $( 0,4 )$ or $( 7 , \infty )$
D) $( - \infty , 4 )$ or $( 7 , \infty )$

Accepted Answer

To solve the inequality, first rearrange it to $\frac{3x}{7-x} - x < 0$. Simplify to $\frac{3x - x(7-x)}{7-x} < 0$, which becomes $\frac{-x^2 + 10x}{7-x} < 0$. The critical points are $x = 0$, $x = 4$, and $x = 7$. Testing intervals, the solution is $(0, 4)$ or $(7, \infty)$.

Quiz 3: Polynomial and Rational Functions

Solve the inequality. - $\frac { 14 } { x - 6 } > \frac { 12 } { x - 1 }$

Solve the inequality. - $\frac { ( x + 10 ) ( x - 2 ) } { x - 1 } \geq 0$

Solve the inequality. - $\frac { 15 x } { 5 - x } \geq 5 x$

Solve the inequality. - $\frac { ( x - 2 ) ^ { 2 } } { x ^ { 2 } - 9 } > 0$

Solve the inequality. - $x + \frac { 30 } { x } < 13$

Solve the inequality. - $\frac { ( x - 3 ) ( x + 3 ) } { x } \leq 0$

Solve the inequality. - $\frac { ( x - 1 ) ( 3 - x ) } { ( x - 2 ) ^ { 2 } } \leq 0$

Solve the inequality. - $\frac { x ^ { 2 } ( x - 11 ) ( x + 2 ) } { ( x - 4 ) ( x + 8 ) } \geq 0$

Solve the inequality. - $\frac { 5 x ^ { 2 } - 4 x - 9 } { x + 5 } \leq 0$

Solve the inequality. - $\frac { 3 x } { 7 - x } < x$

Quiz 3: Polynomial and Rational Functions

Solve the inequality. - 14x−6>12x−1\frac { 14 } { x - 6 } > \frac { 12 } { x - 1 }x−614​>x−112​

Solve the inequality. - (x+10) (x−2) x−1≥0\frac { ( x + 10 ) ( x - 2 ) } { x - 1 } \geq 0x−1(x+10) (x−2) ​≥0

Solve the inequality. - 15x5−x≥5x\frac { 15 x } { 5 - x } \geq 5 x5−x15x​≥5x

Solve the inequality. - (x−2) 2x2−9>0\frac { ( x - 2 ) ^ { 2 } } { x ^ { 2 } - 9 } > 0x2−9(x−2) 2​>0

Solve the inequality. - x+30x<13x + \frac { 30 } { x } < 13x+x30​<13

Solve the inequality. - (x−3) (x+3) x≤0\frac { ( x - 3 ) ( x + 3 ) } { x } \leq 0x(x−3) (x+3) ​≤0

Solve the inequality. - (x−1) (3−x) (x−2) 2≤0\frac { ( x - 1 ) ( 3 - x ) } { ( x - 2 ) ^ { 2 } } \leq 0(x−2) 2(x−1) (3−x) ​≤0

Solve the inequality. - x2(x−11) (x+2) (x−4) (x+8) ≥0\frac { x ^ { 2 } ( x - 11 ) ( x + 2 ) } { ( x - 4 ) ( x + 8 ) } \geq 0(x−4) (x+8) x2(x−11) (x+2) ​≥0

Solve the inequality. - 5x2−4x−9x+5≤0\frac { 5 x ^ { 2 } - 4 x - 9 } { x + 5 } \leq 0x+55x2−4x−9​≤0

Solve the inequality. - 3x7−x<x\frac { 3 x } { 7 - x } < x7−x3x​<x