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

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

Solve for x

$$x=1$$

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}.

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

Subtract x^{2} from both sides.

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

Subtract 2x from both sides.

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

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

$$2x-x^{2}=1$$

Subtract 1 from both sides.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

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.

$$a+b=2 ab=-\left(-1\right)=1$$

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.

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

To find equation solutions, solve x-1=0 and -x+1=0.

$$x=1 x=1$$

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}.

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

Subtract x^{2} from both sides.

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

Subtract 2x from both sides.

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

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

$$2x-x^{2}=1$$

Subtract 1 from both sides.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

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}.

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

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-1\right)\left(-1\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

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

Multiply 4 times -1.

$$x=\frac{-2±\sqrt{4-4}}{2\left(-1\right)}$$

Add 4 to -4.

$$x=\frac{-2±\sqrt{0}}{2\left(-1\right)}$$

Take the square root of 0.

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

Multiply 2 times -1.

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

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}.

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

Subtract x^{2} from both sides.

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

Subtract 2x from both sides.

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

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

$$2x-x^{2}=1$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 2 by -1.

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

Divide 1 by -1.

$$x^{2}-2x=-1$$

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Add -1 to 1.

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

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.

$$x-1=0 x-1=0$$

Add 1 to both sides of the equation.

$$x=1 x=1$$

The equation is now solved. Solutions are the same.

$$x=1$$