Solve 9(x-2)^2=(2x+3)^2

$$9 ( x - 2 ) ^ { 2 } = ( 2 x + 3 ) ^ { 2 }$$

Solve for x

$$x=\frac{3}{5}=0.6$$

$$x=9$$

Steps Using Factoring By Grouping

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

$$9\left(x^{2}-4x+4\right)=\left(2x+3\right)^{2}$$

Use the distributive property to multiply 9 by x^{2}-4x+4.

$$9x^{2}-36x+36=\left(2x+3\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+3\right)^{2}.

$$9x^{2}-36x+36=4x^{2}+12x+9$$

Subtract 4x^{2} from both sides.

$$9x^{2}-36x+36-4x^{2}=12x+9$$

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

$$5x^{2}-36x+36=12x+9$$

Subtract 12x from both sides.

$$5x^{2}-36x+36-12x=9$$

Combine -36x and -12x to get -48x.

$$5x^{2}-48x+36=9$$

Subtract 9 from both sides.

$$5x^{2}-48x+36-9=0$$

Subtract 9 from 36 to get 27.

$$5x^{2}-48x+27=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx+27. To find a and b, set up a system to be solved.

$$a+b=-48 ab=5\times 27=135$$

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 135.

$$-1,-135 -3,-45 -5,-27 -9,-15$$

Calculate the sum for each pair.

$$-1-135=-136 -3-45=-48 -5-27=-32 -9-15=-24$$

The solution is the pair that gives sum -48.

$$a=-45 b=-3$$

Rewrite 5x^{2}-48x+27 as \left(5x^{2}-45x\right)+\left(-3x+27\right).

$$\left(5x^{2}-45x\right)+\left(-3x+27\right)$$

Factor out 5x in the first and -3 in the second group.

$$5x\left(x-9\right)-3\left(x-9\right)$$

Factor out common term x-9 by using distributive property.

$$\left(x-9\right)\left(5x-3\right)$$

To find equation solutions, solve x-9=0 and 5x-3=0.

$$x=9 x=\frac{3}{5}$$

Steps Using the Quadratic Formula

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

$$9\left(x^{2}-4x+4\right)=\left(2x+3\right)^{2}$$

Use the distributive property to multiply 9 by x^{2}-4x+4.

$$9x^{2}-36x+36=\left(2x+3\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+3\right)^{2}.

$$9x^{2}-36x+36=4x^{2}+12x+9$$

Subtract 4x^{2} from both sides.

$$9x^{2}-36x+36-4x^{2}=12x+9$$

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

$$5x^{2}-36x+36=12x+9$$

Subtract 12x from both sides.

$$5x^{2}-36x+36-12x=9$$

Combine -36x and -12x to get -48x.

$$5x^{2}-48x+36=9$$

Subtract 9 from both sides.

$$5x^{2}-48x+36-9=0$$

Subtract 9 from 36 to get 27.

$$5x^{2}-48x+27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -48 for b, and 27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-48\right)±\sqrt{\left(-48\right)^{2}-4\times 5\times 27}}{2\times 5}$$

Square -48.

$$x=\frac{-\left(-48\right)±\sqrt{2304-4\times 5\times 27}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-48\right)±\sqrt{2304-20\times 27}}{2\times 5}$$

Multiply -20 times 27.

$$x=\frac{-\left(-48\right)±\sqrt{2304-540}}{2\times 5}$$

Add 2304 to -540.

$$x=\frac{-\left(-48\right)±\sqrt{1764}}{2\times 5}$$

Take the square root of 1764.

$$x=\frac{-\left(-48\right)±42}{2\times 5}$$

The opposite of -48 is 48.

$$x=\frac{48±42}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{48±42}{10}$$

Now solve the equation x=\frac{48±42}{10} when ± is plus. Add 48 to 42.

$$x=\frac{90}{10}$$

Divide 90 by 10.

$$x=9$$

Now solve the equation x=\frac{48±42}{10} when ± is minus. Subtract 42 from 48.

$$x=\frac{6}{10}$$

Reduce the fraction \frac{6}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{3}{5}$$

The equation is now solved.

$$x=9 x=\frac{3}{5}$$

Steps for Completing the Square

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

$$9\left(x^{2}-4x+4\right)=\left(2x+3\right)^{2}$$

Use the distributive property to multiply 9 by x^{2}-4x+4.

$$9x^{2}-36x+36=\left(2x+3\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+3\right)^{2}.

$$9x^{2}-36x+36=4x^{2}+12x+9$$

Subtract 4x^{2} from both sides.

$$9x^{2}-36x+36-4x^{2}=12x+9$$

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

$$5x^{2}-36x+36=12x+9$$

Subtract 12x from both sides.

$$5x^{2}-36x+36-12x=9$$

Combine -36x and -12x to get -48x.

$$5x^{2}-48x+36=9$$

Subtract 36 from both sides.

$$5x^{2}-48x=9-36$$

Subtract 36 from 9 to get -27.

$$5x^{2}-48x=-27$$

Divide both sides by 5.

$$\frac{5x^{2}-48x}{5}=-\frac{27}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{48}{5}x=-\frac{27}{5}$$

Divide -\frac{48}{5}, the coefficient of the x term, by 2 to get -\frac{24}{5}. Then add the square of -\frac{24}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{48}{5}x+\left(-\frac{24}{5}\right)^{2}=-\frac{27}{5}+\left(-\frac{24}{5}\right)^{2}$$

Square -\frac{24}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{48}{5}x+\frac{576}{25}=-\frac{27}{5}+\frac{576}{25}$$

Add -\frac{27}{5} to \frac{576}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{48}{5}x+\frac{576}{25}=\frac{441}{25}$$

Factor x^{2}-\frac{48}{5}x+\frac{576}{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}.

$$\left(x-\frac{24}{5}\right)^{2}=\frac{441}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{24}{5}\right)^{2}}=\sqrt{\frac{441}{25}}$$

Simplify.

$$x-\frac{24}{5}=\frac{21}{5} x-\frac{24}{5}=-\frac{21}{5}$$

Add \frac{24}{5} to both sides of the equation.

$$x=9 x=\frac{3}{5}$$