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College Algebra : Solving Equations and Inequallities

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Example Question #1 : Solving Non Quadratic Polynomials

Give all real solutions of the following equation:

$x^{4} +5x^{2} - 36 = 0$

Possible Answers:

$\left \{ -3,3\right \}$

$\left \{ -3, -2\right \}$

$\left \{ -3, -2, 2, 3\right \}$

$\left \{ -2,2\right \}$

$\left \{ 2,3\right \}$

Correct answer:

$\left \{ -2,2\right \}$

Explanation:

By substituting $u = x^{2}$ - and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ this can be rewritten as a quadratic equation, and solved as such:

$x^{4} +5x^{2} - 36 = 0$

$u^{2} + 5u - 36 = 0$

We are looking to factor the quadratic expression as (u+?)(u+?) , replacing the two question marks with integers with product and sum 5; these integers are 9,-4 .

(u+9)(u-4) = 0

Substitute back:

$(x^{2}+9)(x^{2}-4) = 0$

The first factor cannot be factored further. The second factor, however, can itself be factored as the difference of squares:

$(x^{2}+9)(x+2)(x-2) = 0$

Set each factor to zero and solve:

$x^{2}+9= 0$

$x^{2}= -9$

Since no real number squared is equal to a negative number, no real solution presents itself here.

$x-2= 0 \Rightarrow x = 2$

$x+2 = 0 \Rightarrow x = - 2$

The solution set is $\left \{ -2, 2\right \}$ .

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Example Question #1 : Solving Equations And Inequallities

Give all real solutions of the following equation:

$x^{4} - 13x^{2} + 36 = 0$

Possible Answers:

$\left \{ 2, 3\right \}$

The equation has no real solutions.

$\left \{ 4,9\right \}$

$\left \{ -9,-4,4,9\right \}$

$\left \{ -3, -2, 2, 3\right \}$

Correct answer:

$\left \{ -3, -2, 2, 3\right \}$

Explanation:

By substituting $u = x^{2}$ - and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ this can be rewritten as a quadratic equation, and solved as such:

$x^{4} - 13x^{2} + 36 = 0$

$u^{2} - 13u + 36 = 0$

We are looking to factor the quadratic expression as (u+?)(u+?) , replacing the two question marks with integers with product 36 and sum ; these integers are -9,-4 .

(u-9)(u-4) = 0

Substitute back:

$(x^{2}-9)(x^{2}-4) = 0$

These factors can themselves be factored as the difference of squares:

(x+3)(x-3)(x+2)(x-2) = 0

Set each factor to zero and solve:

$x-3 = 0 \Rightarrow x = 3$

$x-2= 0 \Rightarrow x = 2$

$x+2 = 0 \Rightarrow x = - 2$

$x+3 = 0 \Rightarrow x = - 3$

The solution set is $\left \{ -3, -2, 2, 3\right \}$ .

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Example Question #1 : Linear Equations

Solve for :

8x-5-4x=-x+10

Possible Answers:

Correct answer:

Explanation:

8x-5-4x=-x+10 can be simplified to become

4x-5=-x+10

Then, you can further simplify by adding 5 and to both sides to get 5x=15 .

Then, you can divide both sides by 5 to get x=3 .

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Example Question #1 : How To Find The Solution To An Equation

Solve for :

6x-1=12x+8-3x

Possible Answers:

Correct answer:

Explanation:

To solve for , you must first combine the 's on the right side of the equation. This will give you $\ 6x-1=9x+8$ .

Then, subtract and from both sides of the equation to get -9=3x .

Finally, divide both sides by to get the solution x=-3 .

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Example Question #11 : Linear / Rational / Variable Equations

Solve the following equation for :

2(x + 5) = -4

Possible Answers:

-4.5

Correct answer:

Explanation:

The first step is to distribute (multiply) the 2 through the parentheses:

2(x+5) = -4

2x + 10 = -4

Then isolate on the left side of the equation. Subtract the 10 from the left and right side.

2x + 10 = -4

2x = -14

Finally, to isolate , divide the left side by 2 so that the 2 cancels out. Then divide by 2 on the right side as well.

2x = -14

x = -7

You can verify this answer by plugging the into the original equation.

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Example Question #4 : Linear Equations

Solve for :

6x-3-4x+5=3(x-1)

Possible Answers:

Correct answer:

Explanation:

Combine like terms on the left side of the equation: 2x+2=3(x-1)

Use the distributive property to simplify the right side of the equation: 2x+2=3x-3

Next, move the 's to one side and the integers to the other side: x=5

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Example Question #5 : Linear Equations

Solve for x: 12x-5(x-2)=45

Possible Answers:

x=5

x=8

$x=\frac{55}{7}$

$x=\frac{7}{55}$

x=7

Correct answer:

x=5

Explanation:

12x-5(x-2)=45

Simplify the parenthesis:

12x-5x+10=45

Combine the terms with x's:

7x+10=45

Combine constants:

$7x=35\rightarrow \mathbf{x=5}$

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Example Question #1 : Solving Equations And Inequallities

Solve the following equation when y is equal to four.

14y-12x=200

Possible Answers:

x=-14

x=144

x=12

x=-12

Correct answer:

x=-12

Explanation:

Solve the following equation when y is equal to four.

14y-12x=200

To solve this equation, we need to plug in 4 for y and solve.

14(4)-12x=200

56-12x=200

-12x=144

x=-12

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Example Question #1 : Linear Equations

Solve the following:

3x-7=2

Possible Answers:

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

x=-3

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

x=3

Correct answer:

x=3

Explanation:

To solve, we must isolate x. In order to do that, we must first add 7 to both sides.

3x-7+7=2+7

3x=9

Next, we must divide both sides by 3.

$\frac{3x}{3}=\frac{9}{3}$

x=3

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Example Question #1 : Linear Equations

Write an equation of the line passing through (5,10) and (10,2).

Possible Answers:

$y=\frac{5}{8}x+9$

None of these.

$y=\frac{8}{5}x-18$

$y= \frac{8}{5}x+18$

$y= -\frac{8}{5}x+18$

Correct answer:

$y= -\frac{8}{5}x+18$

Explanation:

To find this line, first find the slope (m) between the two coordinate points. Then use the point-slope formula to find a line with that same slope passing through a particular point.

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}= \frac{2-10}{10-5}=-\frac{8}{5}$

$y-y_{1}=m(x-x_{1})$

$y-10=-\frac{8}{5}(x-5) \rightarrow y-10=-\frac{8}{5}x+8 \rightarrow \boldsymbol{y=-\frac{8}{5}x+18}$

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College Algebra : Solving Equations and Inequallities

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #1 : Solving Non Quadratic Polynomials

Example Question #1 : Solving Equations And Inequallities

Example Question #1 : Linear Equations

Example Question #1 : How To Find The Solution To An Equation

Example Question #11 : Linear / Rational / Variable Equations

Example Question #4 : Linear Equations

Example Question #5 : Linear Equations

Example Question #1 : Solving Equations And Inequallities

Example Question #1 : Linear Equations

Example Question #1 : Linear Equations

All College Algebra Resources

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