Solve x+2/4-[2(1-x)]^2/64=-(x+1/4)^2+(1-1/2)x+frac{1}{2}

$$\frac { x + 2 } { 4 } - \frac { [ 2 ( 1 - x ) ] ^ { 2 } } { 64 } = - ( \frac { x + 1 } { 4 } ) ^ { 2 } + ( 1 - \frac { 1 } { 2 } ) x + \frac { 1 } { 2 }$$

Solve for x (complex solution)

$$x\in \mathrm{C}$$

Short Solution Steps

Multiply both sides of the equation by 64, the least common multiple of 4,64,2.

$$16\left(x+2\right)-\left(2\left(1-x\right)\right)^{2}=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

Use the distributive property to multiply 16 by x+2.

$$16x+32-\left(2\left(1-x\right)\right)^{2}=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

Use the distributive property to multiply 2 by 1-x.

$$16x+32-\left(2-2x\right)^{2}=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

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

$$16x+32-\left(4-8x+4x^{2}\right)=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

To find the opposite of 4-8x+4x^{2}, find the opposite of each term.

$$16x+32-4+8x-4x^{2}=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

Subtract 4 from 32 to get 28.

$$16x+28+8x-4x^{2}=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

Combine 16x and 8x to get 24x.

$$24x+28-4x^{2}=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

To raise \frac{x+1}{4} to a power, raise both numerator and denominator to the power and then divide.

$$24x+28-4x^{2}=-64\times \frac{\left(x+1\right)^{2}}{4^{2}}+64\left(1-\frac{1}{2}\right)x+32$$

Express -64\times \frac{\left(x+1\right)^{2}}{4^{2}} as a single fraction.

$$24x+28-4x^{2}=\frac{-64\left(x+1\right)^{2}}{4^{2}}+64\left(1-\frac{1}{2}\right)x+32$$

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

$$24x+28-4x^{2}=\frac{-64\left(x+1\right)^{2}}{4^{2}}+64\times \frac{1}{2}x+32$$

Multiply 64 and \frac{1}{2} to get 32.

$$24x+28-4x^{2}=\frac{-64\left(x+1\right)^{2}}{4^{2}}+32x+32$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 32x+32 times \frac{4^{2}}{4^{2}}.

$$24x+28-4x^{2}=\frac{-64\left(x+1\right)^{2}}{4^{2}}+\frac{\left(32x+32\right)\times 4^{2}}{4^{2}}$$

Since \frac{-64\left(x+1\right)^{2}}{4^{2}} and \frac{\left(32x+32\right)\times 4^{2}}{4^{2}} have the same denominator, add them by adding their numerators.

$$24x+28-4x^{2}=\frac{-64\left(x+1\right)^{2}+\left(32x+32\right)\times 4^{2}}{4^{2}}$$

Do the multiplications in -64\left(x+1\right)^{2}+\left(32x+32\right)\times 4^{2}.

$$24x+28-4x^{2}=\frac{-64x^{2}-128x-64+512x+512}{4^{2}}$$

Combine like terms in -64x^{2}-128x-64+512x+512.

$$24x+28-4x^{2}=\frac{-64x^{2}+384x+448}{4^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$24x+28-4x^{2}=\frac{-64x^{2}+384x+448}{16}$$

Divide each term of -64x^{2}+384x+448 by 16 to get -4x^{2}+24x+28.

$$24x+28-4x^{2}=-4x^{2}+24x+28$$

Add 4x^{2} to both sides.

$$24x+28-4x^{2}+4x^{2}=24x+28$$

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

$$24x+28=24x+28$$

Subtract 24x from both sides.

$$24x+28-24x=28$$

Combine 24x and -24x to get 0.

$$28=28$$

Compare 28 and 28.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{C}$$

Solve for x

$$x\in \mathrm{R}$$

Short Solution Steps

Multiply both sides of the equation by 64, the least common multiple of 4,64,2.

$$16\left(x+2\right)-\left(2\left(1-x\right)\right)^{2}=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

Use the distributive property to multiply 16 by x+2.

$$16x+32-\left(2\left(1-x\right)\right)^{2}=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

Use the distributive property to multiply 2 by 1-x.

$$16x+32-\left(2-2x\right)^{2}=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

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

$$16x+32-\left(4-8x+4x^{2}\right)=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

To find the opposite of 4-8x+4x^{2}, find the opposite of each term.

$$16x+32-4+8x-4x^{2}=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

Subtract 4 from 32 to get 28.

$$16x+28+8x-4x^{2}=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

Combine 16x and 8x to get 24x.

$$24x+28-4x^{2}=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

To raise \frac{x+1}{4} to a power, raise both numerator and denominator to the power and then divide.

$$24x+28-4x^{2}=-64\times \frac{\left(x+1\right)^{2}}{4^{2}}+64\left(1-\frac{1}{2}\right)x+32$$

Express -64\times \frac{\left(x+1\right)^{2}}{4^{2}} as a single fraction.

$$24x+28-4x^{2}=\frac{-64\left(x+1\right)^{2}}{4^{2}}+64\left(1-\frac{1}{2}\right)x+32$$

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

$$24x+28-4x^{2}=\frac{-64\left(x+1\right)^{2}}{4^{2}}+64\times \frac{1}{2}x+32$$

Multiply 64 and \frac{1}{2} to get 32.

$$24x+28-4x^{2}=\frac{-64\left(x+1\right)^{2}}{4^{2}}+32x+32$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 32x+32 times \frac{4^{2}}{4^{2}}.

$$24x+28-4x^{2}=\frac{-64\left(x+1\right)^{2}}{4^{2}}+\frac{\left(32x+32\right)\times 4^{2}}{4^{2}}$$

Since \frac{-64\left(x+1\right)^{2}}{4^{2}} and \frac{\left(32x+32\right)\times 4^{2}}{4^{2}} have the same denominator, add them by adding their numerators.

$$24x+28-4x^{2}=\frac{-64\left(x+1\right)^{2}+\left(32x+32\right)\times 4^{2}}{4^{2}}$$

Do the multiplications in -64\left(x+1\right)^{2}+\left(32x+32\right)\times 4^{2}.

$$24x+28-4x^{2}=\frac{-64x^{2}-128x-64+512x+512}{4^{2}}$$

Combine like terms in -64x^{2}-128x-64+512x+512.

$$24x+28-4x^{2}=\frac{-64x^{2}+384x+448}{4^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$24x+28-4x^{2}=\frac{-64x^{2}+384x+448}{16}$$

Divide each term of -64x^{2}+384x+448 by 16 to get -4x^{2}+24x+28.

$$24x+28-4x^{2}=-4x^{2}+24x+28$$

Add 4x^{2} to both sides.

$$24x+28-4x^{2}+4x^{2}=24x+28$$

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

$$24x+28=24x+28$$

Subtract 24x from both sides.

$$24x+28-24x=28$$

Combine 24x and -24x to get 0.

$$28=28$$

Compare 28 and 28.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{R}$$