Intermediate Single-Variable Algebra

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Algebra 2 › Intermediate Single-Variable Algebra

Questions 1 - 10

Josephine wanted to solve the quadratic equation below by completing the square. Her first two steps are shown below:

Equation:

Step 1:

Step 2:

Which of the following equations would best represent the next step in solving the equation?

Explanation

To solve an equation by completing the square, you must factor the perfect square. The factored form of is . Once the left side of the equation is factored, you may take the square root of both sides.

Solve for .

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

Explanation

To solve for the variable , isolate the variable on one side of the equation with all other constants on the other side. To accomplish this perform the opposite operation to manipulate the equation.

First cross multiply.

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

Next, divide by four on both sides.

$\frac{4x}{4}=\frac{12}{4}=\frac{4\cdot 3}{4}=3$

Which value of makes the following expression undefined?

$\frac{x+1}{x-1}$

$\text{All real numbers}$

$\text{Cannot be determined}$

Explanation

A rational expression is undefined when the denominator is zero.

The denominator is zero when .

Which value of makes the following expression undefined?

$\frac{x+1}{x-1}$

$\text{All real numbers}$

$\text{Cannot be determined}$

Explanation

A rational expression is undefined when the denominator is zero.

The denominator is zero when .

Solve: $\frac{2}{3x}+\frac{x}{4}-\frac{1}{x}$

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

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

$\frac{1}{12}x$

$\frac{1}{12x}$

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

Explanation

Determine the least common denominator. Each term will need an x-variable, and all three denominators will need a common coefficient.

The least common denominator is .

Convert the fractions.

$\frac{2(4)}{3x(4)}+\frac{x(3x)}{4(3x)}-\frac{1(12)}{x(12)} = \frac{8}{12x}+\frac{3x^2}{12x}-\frac{12}{12x}$

The expression becomes:

$\frac{3x^2-4}{12x} = \frac{x}{4}-\frac{1}{3x}$

The answer is: $\frac{1}{4}x-\frac{1}{3x}$

Subtract: $\frac{3}{2-x}-\frac{5}{2x}$

$-\frac{11x-10}{2x^2-4x}$

$-\frac{11x-10}{2x^2-2x}$

$-\frac{x+10}{2x^2-8x}$

$-\frac{4x+10}{2x^2-2x}$

$\frac{x+10}{2x^2-2x}$

Explanation

Multiply the denominators to get the least common denominator. We can then convert both fractions so that the denominators are alike.

$\frac{3}{2-x}-\frac{5}{2x}=\frac{3(2x)}{(2-x)(2x)}-\frac{5(2-x)}{2x(2-x)}$

Simplify both the top and the bottom.

$\frac{6x}{-2x^2+4x} - \frac{10-5x}{-2x^2+4x}$

Combine the numerators as one fraction. Be careful with the second fraction since the entire numerator is a quantity, which means we will need to brace with parentheses.

$\frac{6x-(10-5x)}{-2x^2+4x} =\frac{6x-10+5x}{-2x^2+4x} = \frac{11x-10}{-2x^2+4x}$

Pull out a common factor of negative one in the denominator. This allows us to rewrite the fraction with the negative sign in front of the fraction.

$\frac{11x-10}{-2x^2+4x}= \frac{11x-10}{-(2x^2-4x)}= -\frac{11x-10}{2x^2-4x}$

The answer is: $-\frac{11x-10}{2x^2-4x}$

If , , and , what is ?

Explanation

To find , we must start inwards and work our way outwards, i.e. starting with :

We can now use this value to find as follows:

Our final answer is therefore

Solve for :

$\frac{3}{x-1} + \frac{4}{x- 3 } = 5$

$x = 1 \frac{2}{5} \textrm{ or } x = 4$

$x = 1 \frac{2}{5}$

$x = 4\textrm{ or }x = 7$

Explanation

Multiply both sides by :

$\frac{3}{x-1} + \frac{4}{x- 3 } = 5$

$\frac{3}{x-1} \cdot (x-1) (x-3)+ \frac{4}{x- 3 } \cdot (x-1) (x-3) = 5\cdot (x-1) (x-3)$

$3 \cdot (x-3)+4\cdot (x-1) = 5\cdot (x-1) (x-3)$

$3x-9 + 4x-4 = 5 (x ^{2 } -4x+3)$

$7x-13= 5 x ^{2 } -20x+15$

$7x -13 -7x + 13= 5 x ^{2 } -20x -7x +15+ 13$

$5 x ^{2 } -27x + 28 = 0$

Factor this using the -method. We split the middle term using two integers whose sum is and whose product is $5 \cdot28 = 140$ . These integers are :

$5 x ^{2 } -20x -7x + 28 = 0$

$\left (5 x ^{2 } -20x \right ) - \left (7x - 2 8\right ) = 0$

$5x \left (x-4 \right ) - 7 \left (x-4 \right ) = 0$

$\left (5x - 7 \right )\left (x-4 \right ) = 0$

Set each factor equal to 0 and solve separately:

$5x \div 5 = 7 \div 5$

$x = \frac{7}{5} = 1 \frac{2}{5}$

Solve for :

$\frac{3}{x-1} + \frac{4}{x- 3 } = 5$

$x = 1 \frac{2}{5} \textrm{ or } x = 4$

$x = 1 \frac{2}{5}$

$x = 4\textrm{ or }x = 7$

Explanation

Multiply both sides by :

$\frac{3}{x-1} + \frac{4}{x- 3 } = 5$

$\frac{3}{x-1} \cdot (x-1) (x-3)+ \frac{4}{x- 3 } \cdot (x-1) (x-3) = 5\cdot (x-1) (x-3)$

$3 \cdot (x-3)+4\cdot (x-1) = 5\cdot (x-1) (x-3)$

$3x-9 + 4x-4 = 5 (x ^{2 } -4x+3)$

$7x-13= 5 x ^{2 } -20x+15$

$7x -13 -7x + 13= 5 x ^{2 } -20x -7x +15+ 13$

$5 x ^{2 } -27x + 28 = 0$

Factor this using the -method. We split the middle term using two integers whose sum is and whose product is $5 \cdot28 = 140$ . These integers are :

$5 x ^{2 } -20x -7x + 28 = 0$

$\left (5 x ^{2 } -20x \right ) - \left (7x - 2 8\right ) = 0$

$5x \left (x-4 \right ) - 7 \left (x-4 \right ) = 0$

$\left (5x - 7 \right )\left (x-4 \right ) = 0$

Set each factor equal to 0 and solve separately:

$5x \div 5 = 7 \div 5$

$x = \frac{7}{5} = 1 \frac{2}{5}$

Multiply the expressions:

$\small (x^{2} - 2x + 7) (x^{2} - 2x - 7)$

$\small x^{4}-4x^{3}+4x^2-49$

$\small \small x^{4}-4x^{3}+4x^2+49$

$\small \small \small x^{4}-4x^{3}-4x^2-49$

$\small \small \small \small \small x^{4}+4x^{3}-4x^2+49$

$\small \small \small \small x^{4}+4x^{3}-4x^2-49$

Explanation

You can look at this as the sum of two expressions multiplied by the difference of the same two expressions. Use the pattern

$\small (A+B)(A-B) = (A^{2}-B^{2})$ ,

where $\small A = x^{2} - 2x$ and $\small B = 7$ .

$\small \small \small (x^{2} - 2x + 7) (x^{2} - 2x - 7) = (x^{2} - 2x )^{2} - 7^{2} = (x^{2} - 2x )^{2} - 49$

To find $\small (x^{2} - 2x )^{2}$ , you use the formula for perfect squares:

$\small (A-B)^{2} = (A^{2}-2AB+B^{2})$ ,

where $\small \small A = x^{2}$ and $\small B = 2x$ .

$\small \small \small (x^{2}-2x)^{2} = ((x^{2})^{2}-2x^{2}(2x)+(2x)^{2}) = x^{4}-4x^{3}+4x^2$

Substituting above, the final answer is $\small x^{4}-4x^{3}+4x^2-49$ .

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