Solve 1/2x^2-5/6x+4/3=1/12(x-a)(x-b)

$$\frac { 1 } { 2 } x ^ { 2 } - \frac { 5 } { 6 } x + \frac { 4 } { 3 } = \frac { 1 } { 12 } ( x - a ) ( x - b )$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=-\frac{-5x^{2}-bx+10x-16}{b-x}\text{, }&x\neq b\\a\in \mathrm{C}\text{, }&\left(x=\frac{-\sqrt{71}i+5}{6}\text{ and }b=\frac{-\sqrt{71}i+5}{6}\right)\text{ or }\left(x=\frac{5+\sqrt{71}i}{6}\text{ and }b=\frac{5+\sqrt{71}i}{6}\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply \frac{1}{12} by x-a.

$$\frac{1}{2}x^{2}-\frac{5}{6}x+\frac{4}{3}=\left(\frac{1}{12}x-\frac{1}{12}a\right)\left(x-b\right)$$

Use the distributive property to multiply \frac{1}{12}x-\frac{1}{12}a by x-b.

$$\frac{1}{2}x^{2}-\frac{5}{6}x+\frac{4}{3}=\frac{1}{12}x^{2}-\frac{1}{12}xb-\frac{1}{12}ax+\frac{1}{12}ab$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{12}x^{2}-\frac{1}{12}xb-\frac{1}{12}ax+\frac{1}{12}ab=\frac{1}{2}x^{2}-\frac{5}{6}x+\frac{4}{3}$$

Subtract \frac{1}{12}x^{2} from both sides.

$$-\frac{1}{12}xb-\frac{1}{12}ax+\frac{1}{12}ab=\frac{1}{2}x^{2}-\frac{5}{6}x+\frac{4}{3}-\frac{1}{12}x^{2}$$

Combine \frac{1}{2}x^{2} and -\frac{1}{12}x^{2} to get \frac{5}{12}x^{2}.

$$-\frac{1}{12}xb-\frac{1}{12}ax+\frac{1}{12}ab=\frac{5}{12}x^{2}-\frac{5}{6}x+\frac{4}{3}$$

Add \frac{1}{12}xb to both sides.

$$-\frac{1}{12}ax+\frac{1}{12}ab=\frac{5}{12}x^{2}-\frac{5}{6}x+\frac{4}{3}+\frac{1}{12}xb$$

Combine all terms containing a.

$$\left(-\frac{1}{12}x+\frac{1}{12}b\right)a=\frac{5}{12}x^{2}-\frac{5}{6}x+\frac{4}{3}+\frac{1}{12}xb$$

The equation is in standard form.

$$\frac{b-x}{12}a=\frac{bx}{12}+\frac{5x^{2}}{12}-\frac{5x}{6}+\frac{4}{3}$$

Divide both sides by -\frac{1}{12}x+\frac{1}{12}b.

$$\frac{12\times \frac{b-x}{12}a}{b-x}=\frac{12\left(\frac{bx}{12}+\frac{5x^{2}}{12}-\frac{5x}{6}+\frac{4}{3}\right)}{b-x}$$

Dividing by -\frac{1}{12}x+\frac{1}{12}b undoes the multiplication by -\frac{1}{12}x+\frac{1}{12}b.

$$a=\frac{12\left(\frac{bx}{12}+\frac{5x^{2}}{12}-\frac{5x}{6}+\frac{4}{3}\right)}{b-x}$$

Divide \frac{5x^{2}}{12}-\frac{5x}{6}+\frac{4}{3}+\frac{xb}{12} by -\frac{1}{12}x+\frac{1}{12}b.

$$a=\frac{5x^{2}+bx-10x+16}{b-x}$$

Solve for b (complex solution)

$$\left\{\begin{matrix}b=-\frac{-5x^{2}-ax+10x-16}{a-x}\text{, }&x\neq a\\b\in \mathrm{C}\text{, }&\left(x=\frac{-\sqrt{71}i+5}{6}\text{ and }a=\frac{-\sqrt{71}i+5}{6}\right)\text{ or }\left(x=\frac{5+\sqrt{71}i}{6}\text{ and }a=\frac{5+\sqrt{71}i}{6}\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply \frac{1}{12} by x-a.

$$\frac{1}{2}x^{2}-\frac{5}{6}x+\frac{4}{3}=\left(\frac{1}{12}x-\frac{1}{12}a\right)\left(x-b\right)$$

Use the distributive property to multiply \frac{1}{12}x-\frac{1}{12}a by x-b.

$$\frac{1}{2}x^{2}-\frac{5}{6}x+\frac{4}{3}=\frac{1}{12}x^{2}-\frac{1}{12}xb-\frac{1}{12}ax+\frac{1}{12}ba$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{12}x^{2}-\frac{1}{12}xb-\frac{1}{12}ax+\frac{1}{12}ba=\frac{1}{2}x^{2}-\frac{5}{6}x+\frac{4}{3}$$

Subtract \frac{1}{12}x^{2} from both sides.

$$-\frac{1}{12}xb-\frac{1}{12}ax+\frac{1}{12}ba=\frac{1}{2}x^{2}-\frac{5}{6}x+\frac{4}{3}-\frac{1}{12}x^{2}$$

Combine \frac{1}{2}x^{2} and -\frac{1}{12}x^{2} to get \frac{5}{12}x^{2}.

$$-\frac{1}{12}xb-\frac{1}{12}ax+\frac{1}{12}ba=\frac{5}{12}x^{2}-\frac{5}{6}x+\frac{4}{3}$$

Add \frac{1}{12}ax to both sides.

$$-\frac{1}{12}xb+\frac{1}{12}ba=\frac{5}{12}x^{2}-\frac{5}{6}x+\frac{4}{3}+\frac{1}{12}ax$$

Combine all terms containing b.

$$\left(-\frac{1}{12}x+\frac{1}{12}a\right)b=\frac{5}{12}x^{2}-\frac{5}{6}x+\frac{4}{3}+\frac{1}{12}ax$$

The equation is in standard form.

$$\frac{a-x}{12}b=\frac{ax}{12}+\frac{5x^{2}}{12}-\frac{5x}{6}+\frac{4}{3}$$

Divide both sides by -\frac{1}{12}x+\frac{1}{12}a.

$$\frac{12\times \frac{a-x}{12}b}{a-x}=\frac{12\left(\frac{ax}{12}+\frac{5x^{2}}{12}-\frac{5x}{6}+\frac{4}{3}\right)}{a-x}$$

Dividing by -\frac{1}{12}x+\frac{1}{12}a undoes the multiplication by -\frac{1}{12}x+\frac{1}{12}a.

$$b=\frac{12\left(\frac{ax}{12}+\frac{5x^{2}}{12}-\frac{5x}{6}+\frac{4}{3}\right)}{a-x}$$

Divide \frac{5x^{2}}{12}-\frac{5x}{6}+\frac{4}{3}+\frac{ax}{12} by -\frac{1}{12}x+\frac{1}{12}a.

$$b=\frac{5x^{2}+ax-10x+16}{a-x}$$

Solve for a

$$a=-\frac{-5x^{2}-bx+10x-16}{b-x}$$

$$x\neq b$$

Steps for Solving Linear Equation

Use the distributive property to multiply \frac{1}{12} by x-a.

$$\frac{1}{2}x^{2}-\frac{5}{6}x+\frac{4}{3}=\left(\frac{1}{12}x-\frac{1}{12}a\right)\left(x-b\right)$$

Use the distributive property to multiply \frac{1}{12}x-\frac{1}{12}a by x-b.

$$\frac{1}{2}x^{2}-\frac{5}{6}x+\frac{4}{3}=\frac{1}{12}x^{2}-\frac{1}{12}xb-\frac{1}{12}ax+\frac{1}{12}ab$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{12}x^{2}-\frac{1}{12}xb-\frac{1}{12}ax+\frac{1}{12}ab=\frac{1}{2}x^{2}-\frac{5}{6}x+\frac{4}{3}$$

Subtract \frac{1}{12}x^{2} from both sides.

$$-\frac{1}{12}xb-\frac{1}{12}ax+\frac{1}{12}ab=\frac{1}{2}x^{2}-\frac{5}{6}x+\frac{4}{3}-\frac{1}{12}x^{2}$$

Combine \frac{1}{2}x^{2} and -\frac{1}{12}x^{2} to get \frac{5}{12}x^{2}.

$$-\frac{1}{12}xb-\frac{1}{12}ax+\frac{1}{12}ab=\frac{5}{12}x^{2}-\frac{5}{6}x+\frac{4}{3}$$

Add \frac{1}{12}xb to both sides.

$$-\frac{1}{12}ax+\frac{1}{12}ab=\frac{5}{12}x^{2}-\frac{5}{6}x+\frac{4}{3}+\frac{1}{12}xb$$

Combine all terms containing a.

$$\left(-\frac{1}{12}x+\frac{1}{12}b\right)a=\frac{5}{12}x^{2}-\frac{5}{6}x+\frac{4}{3}+\frac{1}{12}xb$$

The equation is in standard form.

$$\frac{b-x}{12}a=\frac{bx}{12}+\frac{5x^{2}}{12}-\frac{5x}{6}+\frac{4}{3}$$

Divide both sides by -\frac{1}{12}x+\frac{1}{12}b.

$$\frac{12\times \frac{b-x}{12}a}{b-x}=\frac{12\left(\frac{bx}{12}+\frac{5x^{2}}{12}-\frac{5x}{6}+\frac{4}{3}\right)}{b-x}$$

Dividing by -\frac{1}{12}x+\frac{1}{12}b undoes the multiplication by -\frac{1}{12}x+\frac{1}{12}b.

$$a=\frac{12\left(\frac{bx}{12}+\frac{5x^{2}}{12}-\frac{5x}{6}+\frac{4}{3}\right)}{b-x}$$

Divide \frac{5x^{2}}{12}-\frac{5x}{6}+\frac{4}{3}+\frac{xb}{12} by -\frac{1}{12}x+\frac{1}{12}b.

$$a=\frac{5x^{2}+bx-10x+16}{b-x}$$

Solve for b

$$b=-\frac{-5x^{2}-ax+10x-16}{a-x}$$

$$x\neq a$$

Steps for Solving Linear Equation

Use the distributive property to multiply \frac{1}{12} by x-a.

$$\frac{1}{2}x^{2}-\frac{5}{6}x+\frac{4}{3}=\left(\frac{1}{12}x-\frac{1}{12}a\right)\left(x-b\right)$$

Use the distributive property to multiply \frac{1}{12}x-\frac{1}{12}a by x-b.

$$\frac{1}{2}x^{2}-\frac{5}{6}x+\frac{4}{3}=\frac{1}{12}x^{2}-\frac{1}{12}xb-\frac{1}{12}ax+\frac{1}{12}ba$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{12}x^{2}-\frac{1}{12}xb-\frac{1}{12}ax+\frac{1}{12}ba=\frac{1}{2}x^{2}-\frac{5}{6}x+\frac{4}{3}$$

Subtract \frac{1}{12}x^{2} from both sides.

$$-\frac{1}{12}xb-\frac{1}{12}ax+\frac{1}{12}ba=\frac{1}{2}x^{2}-\frac{5}{6}x+\frac{4}{3}-\frac{1}{12}x^{2}$$

Combine \frac{1}{2}x^{2} and -\frac{1}{12}x^{2} to get \frac{5}{12}x^{2}.

$$-\frac{1}{12}xb-\frac{1}{12}ax+\frac{1}{12}ba=\frac{5}{12}x^{2}-\frac{5}{6}x+\frac{4}{3}$$

Add \frac{1}{12}ax to both sides.

$$-\frac{1}{12}xb+\frac{1}{12}ba=\frac{5}{12}x^{2}-\frac{5}{6}x+\frac{4}{3}+\frac{1}{12}ax$$

Combine all terms containing b.

$$\left(-\frac{1}{12}x+\frac{1}{12}a\right)b=\frac{5}{12}x^{2}-\frac{5}{6}x+\frac{4}{3}+\frac{1}{12}ax$$

The equation is in standard form.

$$\frac{a-x}{12}b=\frac{ax}{12}+\frac{5x^{2}}{12}-\frac{5x}{6}+\frac{4}{3}$$

Divide both sides by -\frac{1}{12}x+\frac{1}{12}a.

$$\frac{12\times \frac{a-x}{12}b}{a-x}=\frac{12\left(\frac{ax}{12}+\frac{5x^{2}}{12}-\frac{5x}{6}+\frac{4}{3}\right)}{a-x}$$

Dividing by -\frac{1}{12}x+\frac{1}{12}a undoes the multiplication by -\frac{1}{12}x+\frac{1}{12}a.

$$b=\frac{12\left(\frac{ax}{12}+\frac{5x^{2}}{12}-\frac{5x}{6}+\frac{4}{3}\right)}{a-x}$$

Divide \frac{5x^{2}}{12}-\frac{5x}{6}+\frac{4}{3}+\frac{ax}{12} by -\frac{1}{12}x+\frac{1}{12}a.

$$b=\frac{5x^{2}+ax-10x+16}{a-x}$$