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College Algebra : Solutions and Solution Sets

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Example Questions

Example Question #1 : Solving Equations And Inequallities

Give all real solutions of the following equation:

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

We are looking to factor the quadratic expression as $\displaystyle (u+?)(u+?)$, replacing the two question marks with integers with product $\displaystyle -36$ and sum 5; these integers are 9,-4 $\displaystyle 9,-4$.

$\displaystyle (u+9)(u-4) = 0$

Substitute back:

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

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

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

Set each factor to zero and solve:

$x^{2}+9= 0$ $\displaystyle x^{2}+9= 0$

$x^{2}= -9$ $\displaystyle x^{2}= -9$

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

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

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

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

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Example Question #2 : Solving Equations And Inequallities

Give all real solutions of the following equation:

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

Possible Answers:

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

The equation has no real solutions.

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

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

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

Correct answer:

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

Explanation:

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

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

We are looking to factor the quadratic expression as $\displaystyle (u+?)(u+?)$, replacing the two question marks with integers with product 36 and sum $\displaystyle -13$; these integers are -9,-4 $\displaystyle -9,-4$.

$\displaystyle (u-9)(u-4) = 0$

Substitute back:

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

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

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

Set each factor to zero and solve:

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

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

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

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

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

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College Algebra : Solutions and Solution Sets

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #1 : Solving Equations And Inequallities

Example Question #2 : Solving Equations And Inequallities

All College Algebra Resources

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