Algebra II - Math

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Question

$2x + \left | x - 3 \right | = -12$

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Answer

Notice that the equation $2x + \left | x - 3 \right | = -12$ has an term both inside and outside the absolute value expression.

Since the absolute value expression will always produce a positive number and the right side of the equation is negative, a negative number must be added to the result of the absolute value expression to satisfy the equation. Therefore the term outside of the absolute value expression (in this case ) must be negative (meaning must be negative).

Since will be a negative number, the expression within the absolute value will also be negative (before the absolute value is taken). It is thus possible to convert the original equation into an equation that treats the absolute value as a parenthetical expression that will be multiplied by , since any negative value becomes its opposite when taking the absolute value.

$2x + \left | x - 3 \right | = -12$

$\rightarrow 2x+ (x-3)(-1) = -12$

Simplifying and solving this equation for gives the answer:

$\rightarrow 2x -x+3 = -12$

$\rightarrow x + 3 = -12$

$\rightarrow x = -15$

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Question

$2x + \left | x - 3 \right | = -12$

Tap to reveal answer

Answer

Notice that the equation $2x + \left | x - 3 \right | = -12$ has an term both inside and outside the absolute value expression.

Since the absolute value expression will always produce a positive number and the right side of the equation is negative, a negative number must be added to the result of the absolute value expression to satisfy the equation. Therefore the term outside of the absolute value expression (in this case ) must be negative (meaning must be negative).

Since will be a negative number, the expression within the absolute value will also be negative (before the absolute value is taken). It is thus possible to convert the original equation into an equation that treats the absolute value as a parenthetical expression that will be multiplied by , since any negative value becomes its opposite when taking the absolute value.

$2x + \left | x - 3 \right | = -12$

$\rightarrow 2x+ (x-3)(-1) = -12$

Simplifying and solving this equation for gives the answer:

$\rightarrow 2x -x+3 = -12$

$\rightarrow x + 3 = -12$

$\rightarrow x = -15$

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Question

Find the mean and median of the following data set:

$\left \{ 3,4,10,7,6,0,0,3\right\}$

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Answer

The mean is the sum of the numbers divided by their count:

$mean= \frac{3+4+10+7+6+0+0+3}{8}$

To find the median, order the set from smallest to largest:

$\left \{ 0,0,3,3,4,6,7,10 \right \}$

When there are two numbers in the middle, find the average of those numbers:

$median= \frac{3+4}{2}$

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Question

Find the mean and median of the following data set:

$\left \{ 3,4,10,7,6,0,0,3\right\}$

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Answer

The mean is the sum of the numbers divided by their count:

$mean= \frac{3+4+10+7+6+0+0+3}{8}$

To find the median, order the set from smallest to largest:

$\left \{ 0,0,3,3,4,6,7,10 \right \}$

When there are two numbers in the middle, find the average of those numbers:

$median= \frac{3+4}{2}$

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Question

Factor the polynomial completely and solve for .

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Answer

To factor and solve for in the equation

Factor out of the equation

$\rightarrow 8x(4x^2-1)$

Use the "difference of squares" technique to factor the parenthetical term, which provides the completely factored equation:

$\rightarrow 8x(2x+1)(2x-1)$

Any value that causes any one of the three terms , , and to be will be a solution to the equation, therefore

$x = \left \{-\frac{1}{2}, 0, \frac{1}{2} \right \}$

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Question

Solve for :

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Answer

To solve for in the equation

Factor out of the expression on the left of the equation:

$\rightarrow 2(x^2 - 16) =0$

Use the "difference of squares" technique to factor the parenthetical term on the left side of the equation.

$\rightarrow 2(x+4)(x-4)=0$

Any variable that causes any one of the parenthetical terms to become will be a valid solution for the equation. becomes when is , and becomes when is , so the solutions are and .

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Question

Solve for :

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Answer

Pull an out of the left side of the equation.

$\rightarrow x(x^2 - 9) =0$

Use the difference of squares technique to factor the expression in parentheses.

$\rightarrow x(x+3)(x-3) =0$

Any number that causes one of the terms , , or to equal is a solution to the equation. These are , , and , respectively.

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Question

Solve for :

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Answer

To solve for in the equation

Factor out of the expression on the left of the equation:

$\rightarrow 2(x^2 - 16) =0$

Use the "difference of squares" technique to factor the parenthetical term on the left side of the equation.

$\rightarrow 2(x+4)(x-4)=0$

Any variable that causes any one of the parenthetical terms to become will be a valid solution for the equation. becomes when is , and becomes when is , so the solutions are and .

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Question

Solve for :

Tap to reveal answer

Answer

Pull an out of the left side of the equation.

$\rightarrow x(x^2 - 9) =0$

Use the difference of squares technique to factor the expression in parentheses.

$\rightarrow x(x+3)(x-3) =0$

Any number that causes one of the terms , , or to equal is a solution to the equation. These are , , and , respectively.

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Question

Solve for :

Tap to reveal answer

Answer

To solve for in the equation

Factor out of the expression on the left of the equation:

$\rightarrow 2(x^2 - 16) =0$

Use the "difference of squares" technique to factor the parenthetical term on the left side of the equation.

$\rightarrow 2(x+4)(x-4)=0$

Any variable that causes any one of the parenthetical terms to become will be a valid solution for the equation. becomes when is , and becomes when is , so the solutions are and .

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Question

Solve for :

Tap to reveal answer

Answer

Pull an out of the left side of the equation.

$\rightarrow x(x^2 - 9) =0$

Use the difference of squares technique to factor the expression in parentheses.

$\rightarrow x(x+3)(x-3) =0$

Any number that causes one of the terms , , or to equal is a solution to the equation. These are , , and , respectively.

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Question

Solve for :

Tap to reveal answer

Answer

To solve for in the equation

Factor out of the expression on the left of the equation:

$\rightarrow 2(x^2 - 16) =0$

Use the "difference of squares" technique to factor the parenthetical term on the left side of the equation.

$\rightarrow 2(x+4)(x-4)=0$

Any variable that causes any one of the parenthetical terms to become will be a valid solution for the equation. becomes when is , and becomes when is , so the solutions are and .

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Question

Solve for :

Tap to reveal answer

Answer

Pull an out of the left side of the equation.

$\rightarrow x(x^2 - 9) =0$

Use the difference of squares technique to factor the expression in parentheses.

$\rightarrow x(x+3)(x-3) =0$

Any number that causes one of the terms , , or to equal is a solution to the equation. These are , , and , respectively.

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Question

$2x + \left | x - 3 \right | = -12$

Tap to reveal answer

Answer

Notice that the equation $2x + \left | x - 3 \right | = -12$ has an term both inside and outside the absolute value expression.

Since the absolute value expression will always produce a positive number and the right side of the equation is negative, a negative number must be added to the result of the absolute value expression to satisfy the equation. Therefore the term outside of the absolute value expression (in this case ) must be negative (meaning must be negative).

Since will be a negative number, the expression within the absolute value will also be negative (before the absolute value is taken). It is thus possible to convert the original equation into an equation that treats the absolute value as a parenthetical expression that will be multiplied by , since any negative value becomes its opposite when taking the absolute value.

$2x + \left | x - 3 \right | = -12$

$\rightarrow 2x+ (x-3)(-1) = -12$

Simplifying and solving this equation for gives the answer:

$\rightarrow 2x -x+3 = -12$

$\rightarrow x + 3 = -12$

$\rightarrow x = -15$

← Didn't Know|Knew It →

Question

$2x + \left | x - 3 \right | = -12$

Tap to reveal answer

Answer

Notice that the equation $2x + \left | x - 3 \right | = -12$ has an term both inside and outside the absolute value expression.

Since the absolute value expression will always produce a positive number and the right side of the equation is negative, a negative number must be added to the result of the absolute value expression to satisfy the equation. Therefore the term outside of the absolute value expression (in this case ) must be negative (meaning must be negative).

Since will be a negative number, the expression within the absolute value will also be negative (before the absolute value is taken). It is thus possible to convert the original equation into an equation that treats the absolute value as a parenthetical expression that will be multiplied by , since any negative value becomes its opposite when taking the absolute value.

$2x + \left | x - 3 \right | = -12$

$\rightarrow 2x+ (x-3)(-1) = -12$

Simplifying and solving this equation for gives the answer:

$\rightarrow 2x -x+3 = -12$

$\rightarrow x + 3 = -12$

$\rightarrow x = -15$

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Question

Solve for :

$\left | 12x - 39 \right | = 60$

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Answer

Rewrite $\left | 12x - 39 \right | = 60$ as a compound statement and solve each part separately:

$\left | 12x - 39 \right | = 60$

$12x - 39 = - 60 \textrm{ or }12x - 39 = 60$

$12x \div 12 = - 21 \div 12$

$x = -1 \frac{3}{4}$

$12x \div 12 = 99 \div 12$

$x = 8 \frac{1}{4}$

The solution set is $\left \{ -1 \frac{3}{4} ,8 \frac{1}{4}\right \}$

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Question

Solve for :

$\left | 5x + 39 \right | = 44$

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Answer

Rewrite $\left | 5x + 39 \right | = 44$ as a compound statement and solve each part separately:

$\left | 5x + 39 \right | = 44$

$5x + 39 = - 44 \textrm{ or }5x + 39 = 44$

$5x \div 5 = - 83 \div 5$

$x = -16\frac{3}{5}$

$5x \div 5 =5 \div 5$

Therefore the solution set is $\left \{-16\frac{3}{5}, 1 \right \}$ .

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Question

Factor the polynomial completely and solve for .

Tap to reveal answer

Answer

To factor and solve for in the equation

Factor out of the equation

$\rightarrow 8x(4x^2-1)$

Use the "difference of squares" technique to factor the parenthetical term, which provides the completely factored equation:

$\rightarrow 8x(2x+1)(2x-1)$

Any value that causes any one of the three terms , , and to be will be a solution to the equation, therefore

$x = \left \{-\frac{1}{2}, 0, \frac{1}{2} \right \}$

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Question

Solve for :

$\left | 12x - 39 \right | = 60$

Tap to reveal answer

Answer

Rewrite $\left | 12x - 39 \right | = 60$ as a compound statement and solve each part separately:

$\left | 12x - 39 \right | = 60$

$12x - 39 = - 60 \textrm{ or }12x - 39 = 60$

$12x \div 12 = - 21 \div 12$

$x = -1 \frac{3}{4}$

$12x \div 12 = 99 \div 12$

$x = 8 \frac{1}{4}$

The solution set is $\left \{ -1 \frac{3}{4} ,8 \frac{1}{4}\right \}$

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Question

Solve for :

$\left | 5x + 39 \right | = 44$

Tap to reveal answer

Answer

Rewrite $\left | 5x + 39 \right | = 44$ as a compound statement and solve each part separately:

$\left | 5x + 39 \right | = 44$

$5x + 39 = - 44 \textrm{ or }5x + 39 = 44$

$5x \div 5 = - 83 \div 5$

$x = -16\frac{3}{5}$

$5x \div 5 =5 \div 5$

Therefore the solution set is $\left \{-16\frac{3}{5}, 1 \right \}$ .

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