Solve 3(x-1/3)^2+2(x-1/3)-4(1/6x-5/3)>(2x+frac{1}{3})(2x-frac{1}{3})-x^2

$$3 ( x - \frac { 1 } { 3 } ) ^ { 2 } + 2 ( x - \frac { 1 } { 3 } ) - 4 ( \frac { 1 } { 6 } x - \frac { 5 } { 3 } ) > ( 2 x + \frac { 1 } { 3 } ) ( 2 x - \frac { 1 } { 3 } ) - x ^ { 2 }$$

Solve for x

$$x<\frac{29}{3}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\frac{1}{3}\right)^{2}.

$$3\left(x^{2}-\frac{2}{3}x+\frac{1}{9}\right)+2\left(x-\frac{1}{3}\right)-4\left(\frac{1}{6}x-\frac{5}{3}\right)>\left(2x+\frac{1}{3}\right)\left(2x-\frac{1}{3}\right)-x^{2}$$

Use the distributive property to multiply 3 by x^{2}-\frac{2}{3}x+\frac{1}{9}.

$$3x^{2}-2x+\frac{1}{3}+2\left(x-\frac{1}{3}\right)-4\left(\frac{1}{6}x-\frac{5}{3}\right)>\left(2x+\frac{1}{3}\right)\left(2x-\frac{1}{3}\right)-x^{2}$$

Use the distributive property to multiply 2 by x-\frac{1}{3}.

$$3x^{2}-2x+\frac{1}{3}+2x-\frac{2}{3}-4\left(\frac{1}{6}x-\frac{5}{3}\right)>\left(2x+\frac{1}{3}\right)\left(2x-\frac{1}{3}\right)-x^{2}$$

Combine -2x and 2x to get 0.

$$3x^{2}+\frac{1}{3}-\frac{2}{3}-4\left(\frac{1}{6}x-\frac{5}{3}\right)>\left(2x+\frac{1}{3}\right)\left(2x-\frac{1}{3}\right)-x^{2}$$

Subtract \frac{2}{3} from \frac{1}{3} to get -\frac{1}{3}.

$$3x^{2}-\frac{1}{3}-4\left(\frac{1}{6}x-\frac{5}{3}\right)>\left(2x+\frac{1}{3}\right)\left(2x-\frac{1}{3}\right)-x^{2}$$

Use the distributive property to multiply -4 by \frac{1}{6}x-\frac{5}{3}.

$$3x^{2}-\frac{1}{3}-\frac{2}{3}x+\frac{20}{3}>\left(2x+\frac{1}{3}\right)\left(2x-\frac{1}{3}\right)-x^{2}$$

Add -\frac{1}{3} and \frac{20}{3} to get \frac{19}{3}.

$$3x^{2}+\frac{19}{3}-\frac{2}{3}x>\left(2x+\frac{1}{3}\right)\left(2x-\frac{1}{3}\right)-x^{2}$$

Consider \left(2x+\frac{1}{3}\right)\left(2x-\frac{1}{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square \frac{1}{3}.

$$3x^{2}+\frac{19}{3}-\frac{2}{3}x>\left(2x\right)^{2}-\frac{1}{9}-x^{2}$$

Expand \left(2x\right)^{2}.

$$3x^{2}+\frac{19}{3}-\frac{2}{3}x>2^{2}x^{2}-\frac{1}{9}-x^{2}$$

Calculate 2 to the power of 2 and get 4.

$$3x^{2}+\frac{19}{3}-\frac{2}{3}x>4x^{2}-\frac{1}{9}-x^{2}$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}+\frac{19}{3}-\frac{2}{3}x>3x^{2}-\frac{1}{9}$$

Subtract 3x^{2} from both sides.

$$3x^{2}+\frac{19}{3}-\frac{2}{3}x-3x^{2}>-\frac{1}{9}$$

Combine 3x^{2} and -3x^{2} to get 0.

$$\frac{19}{3}-\frac{2}{3}x>-\frac{1}{9}$$

Subtract \frac{19}{3} from both sides.

$$-\frac{2}{3}x>-\frac{1}{9}-\frac{19}{3}$$

Subtract \frac{19}{3} from -\frac{1}{9} to get -\frac{58}{9}.

$$-\frac{2}{3}x>-\frac{58}{9}$$

Multiply both sides by -\frac{3}{2}, the reciprocal of -\frac{2}{3}. Since -\frac{2}{3} is negative, the inequality direction is changed.

$$x<-\frac{58}{9}\left(-\frac{3}{2}\right)$$

Multiply -\frac{58}{9} and -\frac{3}{2} to get \frac{29}{3}.

$$x<\frac{29}{3}$$