Solve left(2x+1right)^2=8x

$${ \left(2x+1 \right) }^{ 2 } =8x$$

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}.

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

Subtract 8x from both sides.

$$4x^{2}+4x+1-8x=0$$

Combine 4x and -8x to get -4x.

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

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.

$$a+b=-4 ab=4\times 1=4$$

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 the first and -1 in the second group.

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

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

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

Rewrite as a binomial square.

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

To find equation solution, solve 2x-1=0.

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

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}.

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

Subtract 8x from both sides.

$$4x^{2}+4x+1-8x=0$$

Combine 4x and -8x to get -4x.

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

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

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

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 4}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-4\right)±\sqrt{16-16}}{2\times 4}$$

Add 16 to -16.

$$x=\frac{-\left(-4\right)±\sqrt{0}}{2\times 4}$$

Take the square root of 0.

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

The opposite of -4 is 4.

$$x=\frac{4}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{4}{8}$$

Reduce the fraction \frac{4}{8} to lowest terms by extracting and canceling out 4.

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

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}.

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

Subtract 8x from both sides.

$$4x^{2}+4x+1-8x=0$$

Combine 4x and -8x to get -4x.

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

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Divide -4 by 4.

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

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.

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

Add \frac{1}{2} to both sides of the equation.

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

The equation is now solved. Solutions are the same.

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