Solve (x+1+frac{sqrt{7}}{2sqrt{2}})(x+1-frac{sqrt{7}}{2sqrt{2}})=x^2+2x+1/8

$$(x+1+ \frac{ \sqrt{ 7 } }{ 2 \sqrt{ 2 } } )(x+1- \frac{ \sqrt{ 7 } }{ 2 \sqrt{ 2 } } )= { x }^{ 2 } +2x+ \frac{ 1 }{ 8 }$$

Solve for x (complex solution)

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

Short Solution Steps

Rationalize the denominator of \frac{\sqrt{7}}{2\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\left(x+1+\frac{\sqrt{7}\sqrt{2}}{2\left(\sqrt{2}\right)^{2}}\right)\left(x+1-\frac{\sqrt{7}}{2\sqrt{2}}\right)=x^{2}+2x+\frac{1}{8}$$

The square of \sqrt{2} is 2.

$$\left(x+1+\frac{\sqrt{7}\sqrt{2}}{2\times 2}\right)\left(x+1-\frac{\sqrt{7}}{2\sqrt{2}}\right)=x^{2}+2x+\frac{1}{8}$$

To multiply \sqrt{7} and \sqrt{2}, multiply the numbers under the square root.

$$\left(x+1+\frac{\sqrt{14}}{2\times 2}\right)\left(x+1-\frac{\sqrt{7}}{2\sqrt{2}}\right)=x^{2}+2x+\frac{1}{8}$$

Multiply 2 and 2 to get 4.

$$\left(x+1+\frac{\sqrt{14}}{4}\right)\left(x+1-\frac{\sqrt{7}}{2\sqrt{2}}\right)=x^{2}+2x+\frac{1}{8}$$

Rationalize the denominator of \frac{\sqrt{7}}{2\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\left(x+1+\frac{\sqrt{14}}{4}\right)\left(x+1-\frac{\sqrt{7}\sqrt{2}}{2\left(\sqrt{2}\right)^{2}}\right)=x^{2}+2x+\frac{1}{8}$$

The square of \sqrt{2} is 2.

$$\left(x+1+\frac{\sqrt{14}}{4}\right)\left(x+1-\frac{\sqrt{7}\sqrt{2}}{2\times 2}\right)=x^{2}+2x+\frac{1}{8}$$

To multiply \sqrt{7} and \sqrt{2}, multiply the numbers under the square root.

$$\left(x+1+\frac{\sqrt{14}}{4}\right)\left(x+1-\frac{\sqrt{14}}{2\times 2}\right)=x^{2}+2x+\frac{1}{8}$$

Multiply 2 and 2 to get 4.

$$\left(x+1+\frac{\sqrt{14}}{4}\right)\left(x+1-\frac{\sqrt{14}}{4}\right)=x^{2}+2x+\frac{1}{8}$$

Use the distributive property to multiply x+1+\frac{\sqrt{14}}{4} by x+1-\frac{\sqrt{14}}{4} and combine like terms.

$$x^{2}+2x+x\left(-\frac{\sqrt{14}}{4}\right)+1+\frac{\sqrt{14}}{4}x+\frac{\sqrt{14}}{4}\left(-\frac{\sqrt{14}}{4}\right)=x^{2}+2x+\frac{1}{8}$$

Express x\left(-\frac{\sqrt{14}}{4}\right) as a single fraction.

$$x^{2}+2x+\frac{-x\sqrt{14}}{4}+1+\frac{\sqrt{14}}{4}x+\frac{\sqrt{14}}{4}\left(-\frac{\sqrt{14}}{4}\right)=x^{2}+2x+\frac{1}{8}$$

Express \frac{\sqrt{14}}{4}x as a single fraction.

$$x^{2}+2x+\frac{-x\sqrt{14}}{4}+1+\frac{\sqrt{14}x}{4}+\frac{\sqrt{14}}{4}\left(-\frac{\sqrt{14}}{4}\right)=x^{2}+2x+\frac{1}{8}$$

Multiply \frac{\sqrt{14}}{4} times -\frac{\sqrt{14}}{4} by multiplying numerator times numerator and denominator times denominator.

$$x^{2}+2x+\frac{-x\sqrt{14}}{4}+1+\frac{\sqrt{14}x}{4}+\frac{-\sqrt{14}\sqrt{14}}{4\times 4}=x^{2}+2x+\frac{1}{8}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2}+2x+1 times \frac{4\times 4}{4\times 4}.

$$\frac{\left(x^{2}+2x+1\right)\times 4\times 4}{4\times 4}+\frac{-x\sqrt{14}}{4}+\frac{\sqrt{14}x}{4}+\frac{-\sqrt{14}\sqrt{14}}{4\times 4}=x^{2}+2x+\frac{1}{8}$$

Since \frac{\left(x^{2}+2x+1\right)\times 4\times 4}{4\times 4} and \frac{-\sqrt{14}\sqrt{14}}{4\times 4} have the same denominator, add them by adding their numerators.

$$\frac{\left(x^{2}+2x+1\right)\times 4\times 4-\sqrt{14}\sqrt{14}}{4\times 4}+\frac{-x\sqrt{14}}{4}+\frac{\sqrt{14}x}{4}=x^{2}+2x+\frac{1}{8}$$

Do the multiplications in \left(x^{2}+2x+1\right)\times 4\times 4-\sqrt{14}\sqrt{14}.

$$\frac{16x^{2}+32x+16-14}{4\times 4}+\frac{-x\sqrt{14}}{4}+\frac{\sqrt{14}x}{4}=x^{2}+2x+\frac{1}{8}$$

Combine like terms in 16x^{2}+32x+16-14.

$$\frac{16x^{2}+32x+2}{4\times 4}+\frac{-x\sqrt{14}}{4}+\frac{\sqrt{14}x}{4}=x^{2}+2x+\frac{1}{8}$$

Factor the expressions that are not already factored in \frac{16x^{2}+32x+2}{4\times 4}.

$$\frac{2\times 8\left(x-\left(-\frac{1}{4}\sqrt{14}-1\right)\right)\left(x-\left(\frac{1}{4}\sqrt{14}-1\right)\right)}{4\times 4}+\frac{-x\sqrt{14}}{4}+\frac{\sqrt{14}x}{4}=x^{2}+2x+\frac{1}{8}$$

Cancel out 2\times 2\times 4 in both numerator and denominator.

$$\left(x-\left(-\frac{1}{4}\sqrt{14}-1\right)\right)\left(x-\left(\frac{1}{4}\sqrt{14}-1\right)\right)+\frac{-x\sqrt{14}}{4}+\frac{\sqrt{14}x}{4}=x^{2}+2x+\frac{1}{8}$$

Expand the expression.

$$x^{2}+2x+\frac{1}{8}+\frac{-x\sqrt{14}}{4}+\frac{\sqrt{14}x}{4}=x^{2}+2x+\frac{1}{8}$$

Since \frac{-x\sqrt{14}}{4} and \frac{\sqrt{14}x}{4} have the same denominator, add them by adding their numerators.

$$x^{2}+2x+\frac{1}{8}+\frac{-x\sqrt{14}+\sqrt{14}x}{4}=x^{2}+2x+\frac{1}{8}$$

Combine like terms in -x\sqrt{14}+\sqrt{14}x.

$$x^{2}+2x+\frac{1}{8}+\frac{0}{4}=x^{2}+2x+\frac{1}{8}$$

Zero divided by any non-zero number gives zero.

$$x^{2}+2x+\frac{1}{8}+0=x^{2}+2x+\frac{1}{8}$$

Add \frac{1}{8} and 0 to get \frac{1}{8}.

$$x^{2}+2x+\frac{1}{8}=x^{2}+2x+\frac{1}{8}$$

Subtract x^{2} from both sides.

$$x^{2}+2x+\frac{1}{8}-x^{2}=2x+\frac{1}{8}$$

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

$$2x+\frac{1}{8}=2x+\frac{1}{8}$$

Subtract 2x from both sides.

$$2x+\frac{1}{8}-2x=\frac{1}{8}$$

Combine 2x and -2x to get 0.

$$\frac{1}{8}=\frac{1}{8}$$

Compare \frac{1}{8} and \frac{1}{8}.

$$\text{true}$$

This is true for any x.

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

Solve for x

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

Short Solution Steps

Rationalize the denominator of \frac{\sqrt{7}}{2\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\left(x+1+\frac{\sqrt{7}\sqrt{2}}{2\left(\sqrt{2}\right)^{2}}\right)\left(x+1-\frac{\sqrt{7}}{2\sqrt{2}}\right)=x^{2}+2x+\frac{1}{8}$$

The square of \sqrt{2} is 2.

$$\left(x+1+\frac{\sqrt{7}\sqrt{2}}{2\times 2}\right)\left(x+1-\frac{\sqrt{7}}{2\sqrt{2}}\right)=x^{2}+2x+\frac{1}{8}$$

To multiply \sqrt{7} and \sqrt{2}, multiply the numbers under the square root.

$$\left(x+1+\frac{\sqrt{14}}{2\times 2}\right)\left(x+1-\frac{\sqrt{7}}{2\sqrt{2}}\right)=x^{2}+2x+\frac{1}{8}$$

Multiply 2 and 2 to get 4.

$$\left(x+1+\frac{\sqrt{14}}{4}\right)\left(x+1-\frac{\sqrt{7}}{2\sqrt{2}}\right)=x^{2}+2x+\frac{1}{8}$$

Rationalize the denominator of \frac{\sqrt{7}}{2\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\left(x+1+\frac{\sqrt{14}}{4}\right)\left(x+1-\frac{\sqrt{7}\sqrt{2}}{2\left(\sqrt{2}\right)^{2}}\right)=x^{2}+2x+\frac{1}{8}$$

The square of \sqrt{2} is 2.

$$\left(x+1+\frac{\sqrt{14}}{4}\right)\left(x+1-\frac{\sqrt{7}\sqrt{2}}{2\times 2}\right)=x^{2}+2x+\frac{1}{8}$$

To multiply \sqrt{7} and \sqrt{2}, multiply the numbers under the square root.

$$\left(x+1+\frac{\sqrt{14}}{4}\right)\left(x+1-\frac{\sqrt{14}}{2\times 2}\right)=x^{2}+2x+\frac{1}{8}$$

Multiply 2 and 2 to get 4.

$$\left(x+1+\frac{\sqrt{14}}{4}\right)\left(x+1-\frac{\sqrt{14}}{4}\right)=x^{2}+2x+\frac{1}{8}$$

Use the distributive property to multiply x+1+\frac{\sqrt{14}}{4} by x+1-\frac{\sqrt{14}}{4} and combine like terms.

$$x^{2}+2x+x\left(-\frac{\sqrt{14}}{4}\right)+1+\frac{\sqrt{14}}{4}x+\frac{\sqrt{14}}{4}\left(-\frac{\sqrt{14}}{4}\right)=x^{2}+2x+\frac{1}{8}$$

Express x\left(-\frac{\sqrt{14}}{4}\right) as a single fraction.

$$x^{2}+2x+\frac{-x\sqrt{14}}{4}+1+\frac{\sqrt{14}}{4}x+\frac{\sqrt{14}}{4}\left(-\frac{\sqrt{14}}{4}\right)=x^{2}+2x+\frac{1}{8}$$

Express \frac{\sqrt{14}}{4}x as a single fraction.

$$x^{2}+2x+\frac{-x\sqrt{14}}{4}+1+\frac{\sqrt{14}x}{4}+\frac{\sqrt{14}}{4}\left(-\frac{\sqrt{14}}{4}\right)=x^{2}+2x+\frac{1}{8}$$

Multiply \frac{\sqrt{14}}{4} times -\frac{\sqrt{14}}{4} by multiplying numerator times numerator and denominator times denominator.

$$x^{2}+2x+\frac{-x\sqrt{14}}{4}+1+\frac{\sqrt{14}x}{4}+\frac{-\sqrt{14}\sqrt{14}}{4\times 4}=x^{2}+2x+\frac{1}{8}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2}+2x+1 times \frac{4\times 4}{4\times 4}.

$$\frac{\left(x^{2}+2x+1\right)\times 4\times 4}{4\times 4}+\frac{-x\sqrt{14}}{4}+\frac{\sqrt{14}x}{4}+\frac{-\sqrt{14}\sqrt{14}}{4\times 4}=x^{2}+2x+\frac{1}{8}$$

Since \frac{\left(x^{2}+2x+1\right)\times 4\times 4}{4\times 4} and \frac{-\sqrt{14}\sqrt{14}}{4\times 4} have the same denominator, add them by adding their numerators.

$$\frac{\left(x^{2}+2x+1\right)\times 4\times 4-\sqrt{14}\sqrt{14}}{4\times 4}+\frac{-x\sqrt{14}}{4}+\frac{\sqrt{14}x}{4}=x^{2}+2x+\frac{1}{8}$$

Do the multiplications in \left(x^{2}+2x+1\right)\times 4\times 4-\sqrt{14}\sqrt{14}.

$$\frac{16x^{2}+32x+16-14}{4\times 4}+\frac{-x\sqrt{14}}{4}+\frac{\sqrt{14}x}{4}=x^{2}+2x+\frac{1}{8}$$

Combine like terms in 16x^{2}+32x+16-14.

$$\frac{16x^{2}+32x+2}{4\times 4}+\frac{-x\sqrt{14}}{4}+\frac{\sqrt{14}x}{4}=x^{2}+2x+\frac{1}{8}$$

Factor the expressions that are not already factored in \frac{16x^{2}+32x+2}{4\times 4}.

$$\frac{2\times 8\left(x-\left(-\frac{1}{4}\sqrt{14}-1\right)\right)\left(x-\left(\frac{1}{4}\sqrt{14}-1\right)\right)}{4\times 4}+\frac{-x\sqrt{14}}{4}+\frac{\sqrt{14}x}{4}=x^{2}+2x+\frac{1}{8}$$

Cancel out 2\times 2\times 4 in both numerator and denominator.

$$\left(x-\left(-\frac{1}{4}\sqrt{14}-1\right)\right)\left(x-\left(\frac{1}{4}\sqrt{14}-1\right)\right)+\frac{-x\sqrt{14}}{4}+\frac{\sqrt{14}x}{4}=x^{2}+2x+\frac{1}{8}$$

Expand the expression.

$$x^{2}+2x+\frac{1}{8}+\frac{-x\sqrt{14}}{4}+\frac{\sqrt{14}x}{4}=x^{2}+2x+\frac{1}{8}$$

Since \frac{-x\sqrt{14}}{4} and \frac{\sqrt{14}x}{4} have the same denominator, add them by adding their numerators.

$$x^{2}+2x+\frac{1}{8}+\frac{-x\sqrt{14}+\sqrt{14}x}{4}=x^{2}+2x+\frac{1}{8}$$

Combine like terms in -x\sqrt{14}+\sqrt{14}x.

$$x^{2}+2x+\frac{1}{8}+\frac{0}{4}=x^{2}+2x+\frac{1}{8}$$

Zero divided by any non-zero number gives zero.

$$x^{2}+2x+\frac{1}{8}+0=x^{2}+2x+\frac{1}{8}$$

Add \frac{1}{8} and 0 to get \frac{1}{8}.

$$x^{2}+2x+\frac{1}{8}=x^{2}+2x+\frac{1}{8}$$

Subtract x^{2} from both sides.

$$x^{2}+2x+\frac{1}{8}-x^{2}=2x+\frac{1}{8}$$

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

$$2x+\frac{1}{8}=2x+\frac{1}{8}$$

Subtract 2x from both sides.

$$2x+\frac{1}{8}-2x=\frac{1}{8}$$

Combine 2x and -2x to get 0.

$$\frac{1}{8}=\frac{1}{8}$$

Compare \frac{1}{8} and \frac{1}{8}.

$$\text{true}$$

This is true for any x.

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