Solve {left(x/2right)}^2+left(x+1right)^2=1

$${ \left( \frac{ x }{ 2 } \right) }^{ 2 } + { \left(x+1 \right) }^{ 2 } =1$$

Solve for x

$$x = -\frac{8}{5} = -1\frac{3}{5} = -1.6$$

$$x=0$$

Steps Using Factoring

To raise \frac{x}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{x^{2}}{2^{2}}+\left(x+1\right)^{2}=1$$

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

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

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

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

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

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

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

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

Combine like terms in x^{2}+4x^{2}+8x+4.

$$\frac{5x^{2}+8x+4}{2^{2}}=1$$

Calculate 2 to the power of 2 and get 4.

$$\frac{5x^{2}+8x+4}{4}=1$$

Divide each term of 5x^{2}+8x+4 by 4 to get \frac{5}{4}x^{2}+2x+1.

$$\frac{5}{4}x^{2}+2x+1=1$$

Subtract 1 from both sides.

$$\frac{5}{4}x^{2}+2x+1-1=0$$

Subtract 1 from 1 to get 0.

$$\frac{5}{4}x^{2}+2x=0$$

Factor out x.

$$x\left(\frac{5}{4}x+2\right)=0$$

To find equation solutions, solve x=0 and \frac{5x}{4}+2=0.

$$x=0 x=-\frac{8}{5}$$

Steps Using the Quadratic Formula

To raise \frac{x}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{x^{2}}{2^{2}}+\left(x+1\right)^{2}=1$$

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

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

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

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

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

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

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

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

Combine like terms in x^{2}+4x^{2}+8x+4.

$$\frac{5x^{2}+8x+4}{2^{2}}=1$$

Calculate 2 to the power of 2 and get 4.

$$\frac{5x^{2}+8x+4}{4}=1$$

Divide each term of 5x^{2}+8x+4 by 4 to get \frac{5}{4}x^{2}+2x+1.

$$\frac{5}{4}x^{2}+2x+1=1$$

Subtract 1 from both sides.

$$\frac{5}{4}x^{2}+2x+1-1=0$$

Subtract 1 from 1 to get 0.

$$\frac{5}{4}x^{2}+2x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{5}{4} for a, 2 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}}}{2\times \frac{5}{4}}$$

Take the square root of 2^{2}.

$$x=\frac{-2±2}{2\times \frac{5}{4}}$$

Multiply 2 times \frac{5}{4}.

$$x=\frac{-2±2}{\frac{5}{2}}$$

Now solve the equation x=\frac{-2±2}{\frac{5}{2}} when ± is plus. Add -2 to 2.

$$x=\frac{0}{\frac{5}{2}}$$

Divide 0 by \frac{5}{2} by multiplying 0 by the reciprocal of \frac{5}{2}.

$$x=0$$

Now solve the equation x=\frac{-2±2}{\frac{5}{2}} when ± is minus. Subtract 2 from -2.

$$x=-\frac{4}{\frac{5}{2}}$$

Divide -4 by \frac{5}{2} by multiplying -4 by the reciprocal of \frac{5}{2}.

$$x=-\frac{8}{5}$$

The equation is now solved.

$$x=0 x=-\frac{8}{5}$$

Steps for Completing the Square

To raise \frac{x}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{x^{2}}{2^{2}}+\left(x+1\right)^{2}=1$$

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

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

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

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

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

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

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

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

Combine like terms in x^{2}+4x^{2}+8x+4.

$$\frac{5x^{2}+8x+4}{2^{2}}=1$$

Calculate 2 to the power of 2 and get 4.

$$\frac{5x^{2}+8x+4}{4}=1$$

Divide each term of 5x^{2}+8x+4 by 4 to get \frac{5}{4}x^{2}+2x+1.

$$\frac{5}{4}x^{2}+2x+1=1$$

Subtract 1 from both sides.

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

Subtract 1 from 1 to get 0.

$$\frac{5}{4}x^{2}+2x=0$$

Divide both sides of the equation by \frac{5}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{5}{4}x^{2}+2x}{\frac{5}{4}}=\frac{0}{\frac{5}{4}}$$

Dividing by \frac{5}{4} undoes the multiplication by \frac{5}{4}.

$$x^{2}+\frac{2}{\frac{5}{4}}x=\frac{0}{\frac{5}{4}}$$

Divide 2 by \frac{5}{4} by multiplying 2 by the reciprocal of \frac{5}{4}.

$$x^{2}+\frac{8}{5}x=\frac{0}{\frac{5}{4}}$$

Divide 0 by \frac{5}{4} by multiplying 0 by the reciprocal of \frac{5}{4}.

$$x^{2}+\frac{8}{5}x=0$$

Divide \frac{8}{5}, the coefficient of the x term, by 2 to get \frac{4}{5}. Then add the square of \frac{4}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{8}{5}x+\left(\frac{4}{5}\right)^{2}=\left(\frac{4}{5}\right)^{2}$$

Square \frac{4}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{8}{5}x+\frac{16}{25}=\frac{16}{25}$$

Factor x^{2}+\frac{8}{5}x+\frac{16}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{4}{5}\right)^{2}=\frac{16}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{4}{5}\right)^{2}}=\sqrt{\frac{16}{25}}$$

Simplify.

$$x+\frac{4}{5}=\frac{4}{5} x+\frac{4}{5}=-\frac{4}{5}$$

Subtract \frac{4}{5} from both sides of the equation.

$$x=0 x=-\frac{8}{5}$$