Solve C=(2x-1/2)^2-(2/3x+2)(2/3x-2)

$$C = ( 2 x - \frac { 1 } { 2 } ) ^ { 2 } - ( \frac { 2 } { 3 } x + 2 ) ( \frac { 2 } { 3 } x - 2 )$$

Solve for C

$$C=\frac{32x^{2}}{9}-2x+\frac{17}{4}$$

Solution Steps

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

$$C=4x^{2}-2x+\frac{1}{4}-\left(\frac{2}{3}x+2\right)\left(\frac{2}{3}x-2\right)$$

Consider \left(\frac{2}{3}x+2\right)\left(\frac{2}{3}x-2\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 2.

$$C=4x^{2}-2x+\frac{1}{4}-\left(\left(\frac{2}{3}x\right)^{2}-4\right)$$

Expand \left(\frac{2}{3}x\right)^{2}.

$$C=4x^{2}-2x+\frac{1}{4}-\left(\left(\frac{2}{3}\right)^{2}x^{2}-4\right)$$

Calculate \frac{2}{3} to the power of 2 and get \frac{4}{9}.

$$C=4x^{2}-2x+\frac{1}{4}-\left(\frac{4}{9}x^{2}-4\right)$$

To find the opposite of \frac{4}{9}x^{2}-4, find the opposite of each term.

$$C=4x^{2}-2x+\frac{1}{4}-\frac{4}{9}x^{2}+4$$

Combine 4x^{2} and -\frac{4}{9}x^{2} to get \frac{32}{9}x^{2}.

$$C=\frac{32}{9}x^{2}-2x+\frac{1}{4}+4$$

Add \frac{1}{4} and 4 to get \frac{17}{4}.

$$C=\frac{32}{9}x^{2}-2x+\frac{17}{4}$$

Solve for x (complex solution)

$$x=\frac{-3\sqrt{32C-127}+9}{32}$$

$$x=\frac{3\sqrt{32C-127}+9}{32}$$

Solve for x

$$x=\frac{-3\sqrt{32C-127}+9}{32}$$

$$x=\frac{3\sqrt{32C-127}+9}{32}\text{, }C\geq \frac{127}{32}$$