Solve =6/3a^{4}-3/3a^{3}+1/3a^2=2a^4-a^3+1/5a^2

$$= \frac { 6 } { 3 } a ^ { 4 } - \frac { 3 } { 3 } a ^ { 3 } + \frac { 1 } { 3 } a ^ { 2 } = 2 a ^ { 4 } - a ^ { 3 } + \frac { 1 } { 5 } a ^ { 2 }$$

Solve for a

$$a=0$$

Steps by Finding Square Root

Divide 6 by 3 to get 2.

$$2a^{4}-\frac{3}{3}a^{3}+\frac{1}{3}a^{2}=2a^{4}-a^{3}+\frac{1}{5}a^{2}$$

Divide 3 by 3 to get 1.

$$2a^{4}-a^{3}+\frac{1}{3}a^{2}=2a^{4}-a^{3}+\frac{1}{5}a^{2}$$

Subtract 2a^{4} from both sides.

$$2a^{4}-a^{3}+\frac{1}{3}a^{2}-2a^{4}=-a^{3}+\frac{1}{5}a^{2}$$

Combine 2a^{4} and -2a^{4} to get 0.

$$-a^{3}+\frac{1}{3}a^{2}=-a^{3}+\frac{1}{5}a^{2}$$

Add a^{3} to both sides.

$$-a^{3}+\frac{1}{3}a^{2}+a^{3}=\frac{1}{5}a^{2}$$

Combine -a^{3} and a^{3} to get 0.

$$\frac{1}{3}a^{2}=\frac{1}{5}a^{2}$$

Subtract \frac{1}{5}a^{2} from both sides.

$$\frac{1}{3}a^{2}-\frac{1}{5}a^{2}=0$$

Combine \frac{1}{3}a^{2} and -\frac{1}{5}a^{2} to get \frac{2}{15}a^{2}.

$$\frac{2}{15}a^{2}=0$$

Multiply both sides by \frac{15}{2}, the reciprocal of \frac{2}{15}. Anything times zero gives zero.

$$a^{2}=0$$

Take the square root of both sides of the equation.

$$a=0 a=0$$

The equation is now solved. Solutions are the same.

$$a=0$$

Steps Using the Quadratic Formula

Divide 6 by 3 to get 2.

$$2a^{4}-\frac{3}{3}a^{3}+\frac{1}{3}a^{2}=2a^{4}-a^{3}+\frac{1}{5}a^{2}$$

Divide 3 by 3 to get 1.

$$2a^{4}-a^{3}+\frac{1}{3}a^{2}=2a^{4}-a^{3}+\frac{1}{5}a^{2}$$

Subtract 2a^{4} from both sides.

$$2a^{4}-a^{3}+\frac{1}{3}a^{2}-2a^{4}=-a^{3}+\frac{1}{5}a^{2}$$

Combine 2a^{4} and -2a^{4} to get 0.

$$-a^{3}+\frac{1}{3}a^{2}=-a^{3}+\frac{1}{5}a^{2}$$

Add a^{3} to both sides.

$$-a^{3}+\frac{1}{3}a^{2}+a^{3}=\frac{1}{5}a^{2}$$

Combine -a^{3} and a^{3} to get 0.

$$\frac{1}{3}a^{2}=\frac{1}{5}a^{2}$$

Subtract \frac{1}{5}a^{2} from both sides.

$$\frac{1}{3}a^{2}-\frac{1}{5}a^{2}=0$$

Combine \frac{1}{3}a^{2} and -\frac{1}{5}a^{2} to get \frac{2}{15}a^{2}.

$$\frac{2}{15}a^{2}=0$$

Multiply both sides by \frac{15}{2}, the reciprocal of \frac{2}{15}. Anything times zero gives zero.

$$a^{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{0±\sqrt{0^{2}}}{2}$$

Take the square root of 0^{2}.

$$a=\frac{0±0}{2}$$

Divide 0 by 2.

$$a=0$$