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

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

Solve for x

$$x=-\frac{1}{2}=-0.5$$

Steps Using Factoring By Grouping

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

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

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

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

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

$$4x^{2}+4x+1=16x^{2}+16x+4$$

Subtract 16x^{2} from both sides.

$$4x^{2}+4x+1-16x^{2}=16x+4$$

Combine 4x^{2} and -16x^{2} to get -12x^{2}.

$$-12x^{2}+4x+1=16x+4$$

Subtract 16x from both sides.

$$-12x^{2}+4x+1-16x=4$$

Combine 4x and -16x to get -12x.

$$-12x^{2}-12x+1=4$$

Subtract 4 from both sides.

$$-12x^{2}-12x+1-4=0$$

Subtract 4 from 1 to get -3.

$$-12x^{2}-12x-3=0$$

Divide both sides by 3.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -4x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=-4 ab=-4\left(-1\right)=4$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -4.

$$a=-2 b=-2$$

Rewrite -4x^{2}-4x-1 as \left(-4x^{2}-2x\right)+\left(-2x-1\right).

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

Factor out 2x in -4x^{2}-2x.

$$2x\left(-2x-1\right)-2x-1$$

Factor out common term -2x-1 by using distributive property.

$$\left(-2x-1\right)\left(2x+1\right)$$

To find equation solutions, solve -2x-1=0 and 2x+1=0.

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

Steps Using the Quadratic Formula

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

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

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

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

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

$$4x^{2}+4x+1=16x^{2}+16x+4$$

Subtract 16x^{2} from both sides.

$$4x^{2}+4x+1-16x^{2}=16x+4$$

Combine 4x^{2} and -16x^{2} to get -12x^{2}.

$$-12x^{2}+4x+1=16x+4$$

Subtract 16x from both sides.

$$-12x^{2}+4x+1-16x=4$$

Combine 4x and -16x to get -12x.

$$-12x^{2}-12x+1=4$$

Subtract 4 from both sides.

$$-12x^{2}-12x+1-4=0$$

Subtract 4 from 1 to get -3.

$$-12x^{2}-12x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -12 for a, -12 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\left(-12\right)\left(-3\right)}}{2\left(-12\right)}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\left(-12\right)\left(-3\right)}}{2\left(-12\right)}$$

Multiply -4 times -12.

$$x=\frac{-\left(-12\right)±\sqrt{144+48\left(-3\right)}}{2\left(-12\right)}$$

Multiply 48 times -3.

$$x=\frac{-\left(-12\right)±\sqrt{144-144}}{2\left(-12\right)}$$

Add 144 to -144.

$$x=\frac{-\left(-12\right)±\sqrt{0}}{2\left(-12\right)}$$

Take the square root of 0.

$$x=-\frac{-12}{2\left(-12\right)}$$

The opposite of -12 is 12.

$$x=\frac{12}{2\left(-12\right)}$$

Multiply 2 times -12.

$$x=\frac{12}{-24}$$

Reduce the fraction \frac{12}{-24} to lowest terms by extracting and canceling out 12.

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

Steps for Completing the Square

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

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

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

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

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

$$4x^{2}+4x+1=16x^{2}+16x+4$$

Subtract 16x^{2} from both sides.

$$4x^{2}+4x+1-16x^{2}=16x+4$$

Combine 4x^{2} and -16x^{2} to get -12x^{2}.

$$-12x^{2}+4x+1=16x+4$$

Subtract 16x from both sides.

$$-12x^{2}+4x+1-16x=4$$

Combine 4x and -16x to get -12x.

$$-12x^{2}-12x+1=4$$

Subtract 1 from both sides.

$$-12x^{2}-12x=4-1$$

Subtract 1 from 4 to get 3.

$$-12x^{2}-12x=3$$

Divide both sides by -12.

$$\frac{-12x^{2}-12x}{-12}=\frac{3}{-12}$$

Dividing by -12 undoes the multiplication by -12.

$$x^{2}+\left(-\frac{12}{-12}\right)x=\frac{3}{-12}$$

Divide -12 by -12.

$$x^{2}+x=\frac{3}{-12}$$

Reduce the fraction \frac{3}{-12} to lowest terms by extracting and canceling out 3.

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

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

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add -\frac{1}{4} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Factor x^{2}+x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{1}{2} from both sides of the equation.

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

The equation is now solved. Solutions are the same.

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