Solve (2x+7)^2=9(x+2)^2

$$( 2 x + 7 ) ^ { 2 } = 9 ( x + 2 ) ^ { 2 }$$

Solve for x

$$x = -\frac{13}{5} = -2\frac{3}{5} = -2.6$$

$$x=1$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+7\right)^{2}.

$$4x^{2}+28x+49=9\left(x+2\right)^{2}$$

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

$$4x^{2}+28x+49=9\left(x^{2}+4x+4\right)$$

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

$$4x^{2}+28x+49=9x^{2}+36x+36$$

Subtract 9x^{2} from both sides.

$$4x^{2}+28x+49-9x^{2}=36x+36$$

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

$$-5x^{2}+28x+49=36x+36$$

Subtract 36x from both sides.

$$-5x^{2}+28x+49-36x=36$$

Combine 28x and -36x to get -8x.

$$-5x^{2}-8x+49=36$$

Subtract 36 from both sides.

$$-5x^{2}-8x+49-36=0$$

Subtract 36 from 49 to get 13.

$$-5x^{2}-8x+13=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -5x^{2}+ax+bx+13. To find a and b, set up a system to be solved.

$$a+b=-8 ab=-5\times 13=-65$$

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 -65.

$$1,-65 5,-13$$

Calculate the sum for each pair.

$$1-65=-64 5-13=-8$$

The solution is the pair that gives sum -8.

$$a=5 b=-13$$

Rewrite -5x^{2}-8x+13 as \left(-5x^{2}+5x\right)+\left(-13x+13\right).

$$\left(-5x^{2}+5x\right)+\left(-13x+13\right)$$

Factor out 5x in the first and 13 in the second group.

$$5x\left(-x+1\right)+13\left(-x+1\right)$$

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

$$\left(-x+1\right)\left(5x+13\right)$$

To find equation solutions, solve -x+1=0 and 5x+13=0.

$$x=1 x=-\frac{13}{5}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+7\right)^{2}.

$$4x^{2}+28x+49=9\left(x+2\right)^{2}$$

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

$$4x^{2}+28x+49=9\left(x^{2}+4x+4\right)$$

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

$$4x^{2}+28x+49=9x^{2}+36x+36$$

Subtract 9x^{2} from both sides.

$$4x^{2}+28x+49-9x^{2}=36x+36$$

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

$$-5x^{2}+28x+49=36x+36$$

Subtract 36x from both sides.

$$-5x^{2}+28x+49-36x=36$$

Combine 28x and -36x to get -8x.

$$-5x^{2}-8x+49=36$$

Subtract 36 from both sides.

$$-5x^{2}-8x+49-36=0$$

Subtract 36 from 49 to get 13.

$$-5x^{2}-8x+13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, -8 for b, and 13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-5\right)\times 13}}{2\left(-5\right)}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\left(-5\right)\times 13}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-\left(-8\right)±\sqrt{64+20\times 13}}{2\left(-5\right)}$$

Multiply 20 times 13.

$$x=\frac{-\left(-8\right)±\sqrt{64+260}}{2\left(-5\right)}$$

Add 64 to 260.

$$x=\frac{-\left(-8\right)±\sqrt{324}}{2\left(-5\right)}$$

Take the square root of 324.

$$x=\frac{-\left(-8\right)±18}{2\left(-5\right)}$$

The opposite of -8 is 8.

$$x=\frac{8±18}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{8±18}{-10}$$

Now solve the equation x=\frac{8±18}{-10} when ± is plus. Add 8 to 18.

$$x=\frac{26}{-10}$$

Reduce the fraction \frac{26}{-10} to lowest terms by extracting and canceling out 2.

$$x=-\frac{13}{5}$$

Now solve the equation x=\frac{8±18}{-10} when ± is minus. Subtract 18 from 8.

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

Divide -10 by -10.

$$x=1$$

The equation is now solved.

$$x=-\frac{13}{5} x=1$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+7\right)^{2}.

$$4x^{2}+28x+49=9\left(x+2\right)^{2}$$

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

$$4x^{2}+28x+49=9\left(x^{2}+4x+4\right)$$

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

$$4x^{2}+28x+49=9x^{2}+36x+36$$

Subtract 9x^{2} from both sides.

$$4x^{2}+28x+49-9x^{2}=36x+36$$

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

$$-5x^{2}+28x+49=36x+36$$

Subtract 36x from both sides.

$$-5x^{2}+28x+49-36x=36$$

Combine 28x and -36x to get -8x.

$$-5x^{2}-8x+49=36$$

Subtract 49 from both sides.

$$-5x^{2}-8x=36-49$$

Subtract 49 from 36 to get -13.

$$-5x^{2}-8x=-13$$

Divide both sides by -5.

$$\frac{-5x^{2}-8x}{-5}=-\frac{13}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\left(-\frac{8}{-5}\right)x=-\frac{13}{-5}$$

Divide -8 by -5.

$$x^{2}+\frac{8}{5}x=-\frac{13}{-5}$$

Divide -13 by -5.

$$x^{2}+\frac{8}{5}x=\frac{13}{5}$$

Divide \frac{8}{5}, the coefficient of the x term, by 2 to get \frac{4}{5}. Then add the square of \frac{4}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{8}{5}x+\left(\frac{4}{5}\right)^{2}=\frac{13}{5}+\left(\frac{4}{5}\right)^{2}$$

Square \frac{4}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{8}{5}x+\frac{16}{25}=\frac{13}{5}+\frac{16}{25}$$

Add \frac{13}{5} to \frac{16}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{8}{5}x+\frac{16}{25}=\frac{81}{25}$$

Factor x^{2}+\frac{8}{5}x+\frac{16}{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}.

$$\left(x+\frac{4}{5}\right)^{2}=\frac{81}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{4}{5}\right)^{2}}=\sqrt{\frac{81}{25}}$$

Simplify.

$$x+\frac{4}{5}=\frac{9}{5} x+\frac{4}{5}=-\frac{9}{5}$$

Subtract \frac{4}{5} from both sides of the equation.

$$x=1 x=-\frac{13}{5}$$