Solve (2x-1)^2=121

$$( 2 x - 1 ) ^ { 2 } = 121$$

Solve for x

$$x=6$$

$$x=-5$$

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}.

$$4x^{2}-4x+1=121$$

Subtract 121 from both sides.

$$4x^{2}-4x+1-121=0$$

Subtract 121 from 1 to get -120.

$$4x^{2}-4x-120=0$$

Divide both sides by 4.

$$x^{2}-x-30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-30. To find a and b, set up a system to be solved.

$$a+b=-1 ab=1\left(-30\right)=-30$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -1.

$$a=-6 b=5$$

Rewrite x^{2}-x-30 as \left(x^{2}-6x\right)+\left(5x-30\right).

$$\left(x^{2}-6x\right)+\left(5x-30\right)$$

Factor out x in the first and 5 in the second group.

$$x\left(x-6\right)+5\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(x+5\right)$$

To find equation solutions, solve x-6=0 and x+5=0.

$$x=6 x=-5$$

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}.

$$4x^{2}-4x+1=121$$

Subtract 121 from both sides.

$$4x^{2}-4x+1-121=0$$

Subtract 121 from 1 to get -120.

$$4x^{2}-4x-120=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -4 for b, and -120 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 4\left(-120\right)}}{2\times 4}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 4\left(-120\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-4\right)±\sqrt{16-16\left(-120\right)}}{2\times 4}$$

Multiply -16 times -120.

$$x=\frac{-\left(-4\right)±\sqrt{16+1920}}{2\times 4}$$

Add 16 to 1920.

$$x=\frac{-\left(-4\right)±\sqrt{1936}}{2\times 4}$$

Take the square root of 1936.

$$x=\frac{-\left(-4\right)±44}{2\times 4}$$

The opposite of -4 is 4.

$$x=\frac{4±44}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{4±44}{8}$$

Now solve the equation x=\frac{4±44}{8} when ± is plus. Add 4 to 44.

$$x=\frac{48}{8}$$

Divide 48 by 8.

$$x=6$$

Now solve the equation x=\frac{4±44}{8} when ± is minus. Subtract 44 from 4.

$$x=-\frac{40}{8}$$

Divide -40 by 8.

$$x=-5$$

The equation is now solved.

$$x=6 x=-5$$

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}.

$$4x^{2}-4x+1=121$$

Subtract 1 from both sides.

$$4x^{2}-4x=121-1$$

Subtract 1 from 121 to get 120.

$$4x^{2}-4x=120$$

Divide both sides by 4.

$$\frac{4x^{2}-4x}{4}=\frac{120}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{4}{4}\right)x=\frac{120}{4}$$

Divide -4 by 4.

$$x^{2}-x=\frac{120}{4}$$

Divide 120 by 4.

$$x^{2}-x=30$$

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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=30+\left(-\frac{1}{2}\right)^{2}$$

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

$$x^{2}-x+\frac{1}{4}=30+\frac{1}{4}$$

Add 30 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{121}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{121}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{121}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{11}{2} x-\frac{1}{2}=-\frac{11}{2}$$

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

$$x=6 x=-5$$