Solve ax^4+5=a

$$a x ^ { 4 } + 5 = a$$

Solve for a (complex solution)

$$a=-\frac{5}{x^{4}-1}$$

$$x\neq -1\text{ and }x\neq -i\text{ and }x\neq i\text{ and }x\neq 1$$

Steps for Solving Linear Equation

Subtract a from both sides.

$$ax^{4}+5-a=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$ax^{4}-a=-5$$

Combine all terms containing a.

$$\left(x^{4}-1\right)a=-5$$

Divide both sides by x^{4}-1.

$$\frac{\left(x^{4}-1\right)a}{x^{4}-1}=-\frac{5}{x^{4}-1}$$

Dividing by x^{4}-1 undoes the multiplication by x^{4}-1.

$$a=-\frac{5}{x^{4}-1}$$

Solve for a

$$a=-\frac{5}{x^{4}-1}$$

$$|x|\neq 1$$

Steps for Solving Linear Equation

Subtract a from both sides.

$$ax^{4}+5-a=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$ax^{4}-a=-5$$

Combine all terms containing a.

$$\left(x^{4}-1\right)a=-5$$

Divide both sides by x^{4}-1.

$$\frac{\left(x^{4}-1\right)a}{x^{4}-1}=-\frac{5}{x^{4}-1}$$

Dividing by x^{4}-1 undoes the multiplication by x^{4}-1.

$$a=-\frac{5}{x^{4}-1}$$

Solve for x (complex solution)

$$x=i\sqrt[4]{\frac{a-5}{a}}$$

$$x=\sqrt[4]{\frac{a-5}{a}}$$

$$x=-\sqrt[4]{\frac{a-5}{a}}$$

$$x=-i\sqrt[4]{\frac{a-5}{a}}\text{, }a\neq 0$$

Solve for x

$$x=\sqrt[4]{\frac{a-5}{a}}$$

$$x=-\sqrt[4]{\frac{a-5}{a}}\text{, }a<0\text{ or }a\geq 5$$