Solve 9x^2-left(2x-1right)^2=0

$$9 { x }^{ 2 } - { \left(2x-1 \right) }^{ 2 } =0$$

Solve for x

$$x=-1$$

$$x=\frac{1}{5}=0.2$$

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}.

$$9x^{2}-\left(4x^{2}-4x+1\right)=0$$

To find the opposite of 4x^{2}-4x+1, find the opposite of each term.

$$9x^{2}-4x^{2}+4x-1=0$$

Combine 9x^{2} and -4x^{2} to get 5x^{2}.

$$5x^{2}+4x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=4 ab=5\left(-1\right)=-5$$

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. The only such pair is the system solution.

$$a=-1 b=5$$

Rewrite 5x^{2}+4x-1 as \left(5x^{2}-x\right)+\left(5x-1\right).

$$\left(5x^{2}-x\right)+\left(5x-1\right)$$

Factor out x in 5x^{2}-x.

$$x\left(5x-1\right)+5x-1$$

Factor out common term 5x-1 by using distributive property.

$$\left(5x-1\right)\left(x+1\right)$$

To find equation solutions, solve 5x-1=0 and x+1=0.

$$x=\frac{1}{5} x=-1$$

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}.

$$9x^{2}-\left(4x^{2}-4x+1\right)=0$$

To find the opposite of 4x^{2}-4x+1, find the opposite of each term.

$$9x^{2}-4x^{2}+4x-1=0$$

Combine 9x^{2} and -4x^{2} to get 5x^{2}.

$$5x^{2}+4x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 4 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 5\left(-1\right)}}{2\times 5}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 5\left(-1\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-4±\sqrt{16-20\left(-1\right)}}{2\times 5}$$

Multiply -20 times -1.

$$x=\frac{-4±\sqrt{16+20}}{2\times 5}$$

Add 16 to 20.

$$x=\frac{-4±\sqrt{36}}{2\times 5}$$

Take the square root of 36.

$$x=\frac{-4±6}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-4±6}{10}$$

Now solve the equation x=\frac{-4±6}{10} when ± is plus. Add -4 to 6.

$$x=\frac{2}{10}$$

Reduce the fraction \frac{2}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{5}$$

Now solve the equation x=\frac{-4±6}{10} when ± is minus. Subtract 6 from -4.

$$x=-\frac{10}{10}$$

Divide -10 by 10.

$$x=-1$$

The equation is now solved.

$$x=\frac{1}{5} x=-1$$

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}.

$$9x^{2}-\left(4x^{2}-4x+1\right)=0$$

To find the opposite of 4x^{2}-4x+1, find the opposite of each term.

$$9x^{2}-4x^{2}+4x-1=0$$

Combine 9x^{2} and -4x^{2} to get 5x^{2}.

$$5x^{2}+4x-1=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$5x^{2}+4x=1$$

Divide both sides by 5.

$$\frac{5x^{2}+4x}{5}=\frac{1}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\frac{4}{5}x=\frac{1}{5}$$

Divide \frac{4}{5}, the coefficient of the x term, by 2 to get \frac{2}{5}. Then add the square of \frac{2}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{4}{5}x+\left(\frac{2}{5}\right)^{2}=\frac{1}{5}+\left(\frac{2}{5}\right)^{2}$$

Square \frac{2}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{4}{5}x+\frac{4}{25}=\frac{1}{5}+\frac{4}{25}$$

Add \frac{1}{5} to \frac{4}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{4}{5}x+\frac{4}{25}=\frac{9}{25}$$

Factor x^{2}+\frac{4}{5}x+\frac{4}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{2}{5}\right)^{2}=\frac{9}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{2}{5}\right)^{2}}=\sqrt{\frac{9}{25}}$$

Simplify.

$$x+\frac{2}{5}=\frac{3}{5} x+\frac{2}{5}=-\frac{3}{5}$$

Subtract \frac{2}{5} from both sides of the equation.

$$x=\frac{1}{5} x=-1$$