Solve left(x-1right)^2+4x=0

$${ \left(x-1 \right) }^{ 2 } +4x=0$$

Solve for x

$$x=-1$$

Steps Using Factoring

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

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

Combine -2x and 4x to get 2x.

$$x^{2}+2x+1=0$$

To solve the equation, factor x^{2}+2x+1 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=2 ab=1$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=1$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x+1\right)\left(x+1\right)$$

Rewrite as a binomial square.

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

To find equation solution, solve x+1=0.

$$x=-1$$

Steps Using Factoring By Grouping

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

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

Combine -2x and 4x to get 2x.

$$x^{2}+2x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\times 1=1$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=1$$

Rewrite x^{2}+2x+1 as \left(x^{2}+x\right)+\left(x+1\right).

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

Factor out x in x^{2}+x.

$$x\left(x+1\right)+x+1$$

Factor out common term x+1 by using distributive property.

$$\left(x+1\right)\left(x+1\right)$$

Rewrite as a binomial square.

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

To find equation solution, solve x+1=0.

$$x=-1$$

Steps Using the Quadratic Formula

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

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

Combine -2x and 4x to get 2x.

$$x^{2}+2x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4}}{2}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4}}{2}$$

Add 4 to -4.

$$x=\frac{-2±\sqrt{0}}{2}$$

Take the square root of 0.

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

Divide -2 by 2.

$$x=-1$$

Steps for Completing the Square

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

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

Combine -2x and 4x to get 2x.

$$x^{2}+2x+1=0$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{0}$$

Simplify.

$$x+1=0 x+1=0$$

Subtract 1 from both sides of the equation.

$$x=-1 x=-1$$

The equation is now solved. Solutions are the same.

$$x=-1$$