Solve left(x-1right)^2+x^2/4=1

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

Solve for x

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

$$x=0$$

Steps Using Factoring

Multiply both sides of the equation by 4.

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

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

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

Use the distributive property to multiply 4 by x^{2}-2x+1.

$$4x^{2}-8x+4+x^{2}=4$$

Combine 4x^{2} and x^{2} to get 5x^{2}.

$$5x^{2}-8x+4=4$$

Subtract 4 from both sides.

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

Subtract 4 from 4 to get 0.

$$5x^{2}-8x=0$$

Factor out x.

$$x\left(5x-8\right)=0$$

To find equation solutions, solve x=0 and 5x-8=0.

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

Steps Using the Quadratic Formula

Multiply both sides of the equation by 4.

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

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

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

Use the distributive property to multiply 4 by x^{2}-2x+1.

$$4x^{2}-8x+4+x^{2}=4$$

Combine 4x^{2} and x^{2} to get 5x^{2}.

$$5x^{2}-8x+4=4$$

Subtract 4 from both sides.

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

Subtract 4 from 4 to get 0.

$$5x^{2}-8x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -8 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}}}{2\times 5}$$

Take the square root of \left(-8\right)^{2}.

$$x=\frac{-\left(-8\right)±8}{2\times 5}$$

The opposite of -8 is 8.

$$x=\frac{8±8}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{8±8}{10}$$

Now solve the equation x=\frac{8±8}{10} when ± is plus. Add 8 to 8.

$$x=\frac{16}{10}$$

Reduce the fraction \frac{16}{10} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{8±8}{10} when ± is minus. Subtract 8 from 8.

$$x=\frac{0}{10}$$

Divide 0 by 10.

$$x=0$$

The equation is now solved.

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

Steps for Completing the Square

Multiply both sides of the equation by 4.

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

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

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

Use the distributive property to multiply 4 by x^{2}-2x+1.

$$4x^{2}-8x+4+x^{2}=4$$

Combine 4x^{2} and x^{2} to get 5x^{2}.

$$5x^{2}-8x+4=4$$

Subtract 4 from both sides.

$$5x^{2}-8x=4-4$$

Subtract 4 from 4 to get 0.

$$5x^{2}-8x=0$$

Divide both sides by 5.

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

Dividing by 5 undoes the multiplication by 5.

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

Divide 0 by 5.

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

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.

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

Add \frac{4}{5} to both sides of the equation.

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