College Algebra - Solutions and Solution Sets

Give all real solutions of the following equation:

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

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

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

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

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

$\left { 2,3\right }$

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:

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

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

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

Give all real solutions of the following equation:

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

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

$\left { 2, 3\right }$

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

$\left { 4,9\right }$

The equation has no real solutions.

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: