Solve x^4-(a+b)x^2+ab=0

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

Solve for a (complex solution)

$$\left\{\begin{matrix}\\a=x^{2}\text{, }&\text{unconditionally}\\a\in \mathrm{C}\text{, }&b=x^{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply a+b by x^{2}.

$$x^{4}-\left(ax^{2}+bx^{2}\right)+ab=0$$

To find the opposite of ax^{2}+bx^{2}, find the opposite of each term.

$$x^{4}-ax^{2}-bx^{2}+ab=0$$

Subtract x^{4} from both sides. Anything subtracted from zero gives its negation.

$$-ax^{2}-bx^{2}+ab=-x^{4}$$

Add bx^{2} to both sides.

$$-ax^{2}+ab=-x^{4}+bx^{2}$$

Combine all terms containing a.

$$\left(-x^{2}+b\right)a=-x^{4}+bx^{2}$$

The equation is in standard form.

$$\left(b-x^{2}\right)a=bx^{2}-x^{4}$$

Divide both sides by -x^{2}+b.

$$\frac{\left(b-x^{2}\right)a}{b-x^{2}}=\frac{x^{2}\left(b-x^{2}\right)}{b-x^{2}}$$

Dividing by -x^{2}+b undoes the multiplication by -x^{2}+b.

$$a=\frac{x^{2}\left(b-x^{2}\right)}{b-x^{2}}$$

Divide \left(-x^{2}+b\right)x^{2} by -x^{2}+b.

$$a=x^{2}$$

Solve for b (complex solution)

$$\left\{\begin{matrix}\\b=x^{2}\text{, }&\text{unconditionally}\\b\in \mathrm{C}\text{, }&a=x^{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply a+b by x^{2}.

$$x^{4}-\left(ax^{2}+bx^{2}\right)+ab=0$$

To find the opposite of ax^{2}+bx^{2}, find the opposite of each term.

$$x^{4}-ax^{2}-bx^{2}+ab=0$$

Subtract x^{4} from both sides. Anything subtracted from zero gives its negation.

$$-ax^{2}-bx^{2}+ab=-x^{4}$$

Add ax^{2} to both sides.

$$-bx^{2}+ab=-x^{4}+ax^{2}$$

Combine all terms containing b.

$$\left(-x^{2}+a\right)b=-x^{4}+ax^{2}$$

The equation is in standard form.

$$\left(a-x^{2}\right)b=ax^{2}-x^{4}$$

Divide both sides by a-x^{2}.

$$\frac{\left(a-x^{2}\right)b}{a-x^{2}}=\frac{x^{2}\left(a-x^{2}\right)}{a-x^{2}}$$

Dividing by a-x^{2} undoes the multiplication by a-x^{2}.

$$b=\frac{x^{2}\left(a-x^{2}\right)}{a-x^{2}}$$

Divide \left(-x^{2}+a\right)x^{2} by a-x^{2}.

$$b=x^{2}$$

Solve for a

$$\left\{\begin{matrix}\\a=x^{2}\text{, }&\text{unconditionally}\\a\in \mathrm{R}\text{, }&b=x^{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply a+b by x^{2}.

$$x^{4}-\left(ax^{2}+bx^{2}\right)+ab=0$$

To find the opposite of ax^{2}+bx^{2}, find the opposite of each term.

$$x^{4}-ax^{2}-bx^{2}+ab=0$$

Subtract x^{4} from both sides. Anything subtracted from zero gives its negation.

$$-ax^{2}-bx^{2}+ab=-x^{4}$$

Add bx^{2} to both sides.

$$-ax^{2}+ab=-x^{4}+bx^{2}$$

Combine all terms containing a.

$$\left(-x^{2}+b\right)a=-x^{4}+bx^{2}$$

The equation is in standard form.

$$\left(b-x^{2}\right)a=bx^{2}-x^{4}$$

Divide both sides by -x^{2}+b.

$$\frac{\left(b-x^{2}\right)a}{b-x^{2}}=\frac{x^{2}\left(b-x^{2}\right)}{b-x^{2}}$$

Dividing by -x^{2}+b undoes the multiplication by -x^{2}+b.

$$a=\frac{x^{2}\left(b-x^{2}\right)}{b-x^{2}}$$

Divide \left(-x^{2}+b\right)x^{2} by -x^{2}+b.

$$a=x^{2}$$

Solve for b

$$\left\{\begin{matrix}\\b=x^{2}\text{, }&\text{unconditionally}\\b\in \mathrm{R}\text{, }&a=x^{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply a+b by x^{2}.

$$x^{4}-\left(ax^{2}+bx^{2}\right)+ab=0$$

To find the opposite of ax^{2}+bx^{2}, find the opposite of each term.

$$x^{4}-ax^{2}-bx^{2}+ab=0$$

Subtract x^{4} from both sides. Anything subtracted from zero gives its negation.

$$-ax^{2}-bx^{2}+ab=-x^{4}$$

Add ax^{2} to both sides.

$$-bx^{2}+ab=-x^{4}+ax^{2}$$

Combine all terms containing b.

$$\left(-x^{2}+a\right)b=-x^{4}+ax^{2}$$

The equation is in standard form.

$$\left(a-x^{2}\right)b=ax^{2}-x^{4}$$

Divide both sides by a-x^{2}.

$$\frac{\left(a-x^{2}\right)b}{a-x^{2}}=\frac{x^{2}\left(a-x^{2}\right)}{a-x^{2}}$$

Dividing by a-x^{2} undoes the multiplication by a-x^{2}.

$$b=\frac{x^{2}\left(a-x^{2}\right)}{a-x^{2}}$$

Divide \left(-x^{2}+a\right)x^{2} by a-x^{2}.

$$b=x^{2}$$