Solutions and Solution Sets - College Algebra

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Question

Give all real solutions of the following equation:

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

Answer

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 , replacing the two question marks with integers with product 36 and sum ; these integers are .

Substitute back:

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

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

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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Question

Give all real solutions of the following equation:

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

Answer

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 , replacing the two question marks with integers with product and sum 5; these integers are .

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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