Solve (2x+1)^2=sqrt{16}

$$( 2 x + 1 ) ^ { 2 } = \sqrt { 16 }$$

Solve for x

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

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

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}.

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

Calculate the square root of 16 and get 4.

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

Subtract 4 from both sides.

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

Subtract 4 from 1 to get -3.

$$4x^{2}+4x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=4 ab=4\left(-3\right)=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 4.

$$a=-2 b=6$$

Rewrite 4x^{2}+4x-3 as \left(4x^{2}-2x\right)+\left(6x-3\right).

$$\left(4x^{2}-2x\right)+\left(6x-3\right)$$

Factor out 2x in the first and 3 in the second group.

$$2x\left(2x-1\right)+3\left(2x-1\right)$$

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

$$\left(2x-1\right)\left(2x+3\right)$$

To find equation solutions, solve 2x-1=0 and 2x+3=0.

$$x=\frac{1}{2} x=-\frac{3}{2}$$

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}.

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

Calculate the square root of 16 and get 4.

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

Subtract 4 from both sides.

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

Subtract 4 from 1 to get -3.

$$4x^{2}+4x-3=0$$

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

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 4\left(-3\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-4±\sqrt{16-16\left(-3\right)}}{2\times 4}$$

Multiply -16 times -3.

$$x=\frac{-4±\sqrt{16+48}}{2\times 4}$$

Add 16 to 48.

$$x=\frac{-4±\sqrt{64}}{2\times 4}$$

Take the square root of 64.

$$x=\frac{-4±8}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-4±8}{8}$$

Now solve the equation x=\frac{-4±8}{8} when ± is plus. Add -4 to 8.

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

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

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

Now solve the equation x=\frac{-4±8}{8} when ± is minus. Subtract 8 from -4.

$$x=-\frac{12}{8}$$

Reduce the fraction \frac{-12}{8} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

$$x=\frac{1}{2} x=-\frac{3}{2}$$

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}.

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

Calculate the square root of 16 and get 4.

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

Subtract 1 from both sides.

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

Subtract 1 from 4 to get 3.

$$4x^{2}+4x=3$$

Divide both sides by 4.

$$\frac{4x^{2}+4x}{4}=\frac{3}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{4}{4}x=\frac{3}{4}$$

Divide 4 by 4.

$$x^{2}+x=\frac{3}{4}$$

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=\frac{3}{4}+\left(\frac{1}{2}\right)^{2}$$

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

$$x^{2}+x+\frac{1}{4}=\frac{3+1}{4}$$

Add \frac{3}{4} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

$$x=\frac{1}{2} x=-\frac{3}{2}$$