Solve 4(x+3)^2=9(2x-1)^2

$$4 ( x + 3 ) ^ { 2 } = 9 ( 2 x - 1 ) ^ { 2 }$$

Solve for x

$$x = \frac{9}{4} = 2\frac{1}{4} = 2.25$$

$$x=-\frac{3}{8}=-0.375$$

Steps Using Factoring By Grouping

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

$$4\left(x^{2}+6x+9\right)=9\left(2x-1\right)^{2}$$

Use the distributive property to multiply 4 by x^{2}+6x+9.

$$4x^{2}+24x+36=9\left(2x-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-1\right)^{2}.

$$4x^{2}+24x+36=9\left(4x^{2}-4x+1\right)$$

Use the distributive property to multiply 9 by 4x^{2}-4x+1.

$$4x^{2}+24x+36=36x^{2}-36x+9$$

Subtract 36x^{2} from both sides.

$$4x^{2}+24x+36-36x^{2}=-36x+9$$

Combine 4x^{2} and -36x^{2} to get -32x^{2}.

$$-32x^{2}+24x+36=-36x+9$$

Add 36x to both sides.

$$-32x^{2}+24x+36+36x=9$$

Combine 24x and 36x to get 60x.

$$-32x^{2}+60x+36=9$$

Subtract 9 from both sides.

$$-32x^{2}+60x+36-9=0$$

Subtract 9 from 36 to get 27.

$$-32x^{2}+60x+27=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -32x^{2}+ax+bx+27. To find a and b, set up a system to be solved.

$$a+b=60 ab=-32\times 27=-864$$

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 -864.

$$-1,864 -2,432 -3,288 -4,216 -6,144 -8,108 -9,96 -12,72 -16,54 -18,48 -24,36 -27,32$$

Calculate the sum for each pair.

$$-1+864=863 -2+432=430 -3+288=285 -4+216=212 -6+144=138 -8+108=100 -9+96=87 -12+72=60 -16+54=38 -18+48=30 -24+36=12 -27+32=5$$

The solution is the pair that gives sum 60.

$$a=72 b=-12$$

Rewrite -32x^{2}+60x+27 as \left(-32x^{2}+72x\right)+\left(-12x+27\right).

$$\left(-32x^{2}+72x\right)+\left(-12x+27\right)$$

Factor out -8x in the first and -3 in the second group.

$$-8x\left(4x-9\right)-3\left(4x-9\right)$$

Factor out common term 4x-9 by using distributive property.

$$\left(4x-9\right)\left(-8x-3\right)$$

To find equation solutions, solve 4x-9=0 and -8x-3=0.

$$x=\frac{9}{4} x=-\frac{3}{8}$$

Steps Using the Quadratic Formula

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

$$4\left(x^{2}+6x+9\right)=9\left(2x-1\right)^{2}$$

Use the distributive property to multiply 4 by x^{2}+6x+9.

$$4x^{2}+24x+36=9\left(2x-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-1\right)^{2}.

$$4x^{2}+24x+36=9\left(4x^{2}-4x+1\right)$$

Use the distributive property to multiply 9 by 4x^{2}-4x+1.

$$4x^{2}+24x+36=36x^{2}-36x+9$$

Subtract 36x^{2} from both sides.

$$4x^{2}+24x+36-36x^{2}=-36x+9$$

Combine 4x^{2} and -36x^{2} to get -32x^{2}.

$$-32x^{2}+24x+36=-36x+9$$

Add 36x to both sides.

$$-32x^{2}+24x+36+36x=9$$

Combine 24x and 36x to get 60x.

$$-32x^{2}+60x+36=9$$

Subtract 9 from both sides.

$$-32x^{2}+60x+36-9=0$$

Subtract 9 from 36 to get 27.

$$-32x^{2}+60x+27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -32 for a, 60 for b, and 27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-60±\sqrt{60^{2}-4\left(-32\right)\times 27}}{2\left(-32\right)}$$

Square 60.

$$x=\frac{-60±\sqrt{3600-4\left(-32\right)\times 27}}{2\left(-32\right)}$$

Multiply -4 times -32.

$$x=\frac{-60±\sqrt{3600+128\times 27}}{2\left(-32\right)}$$

Multiply 128 times 27.

$$x=\frac{-60±\sqrt{3600+3456}}{2\left(-32\right)}$$

Add 3600 to 3456.

$$x=\frac{-60±\sqrt{7056}}{2\left(-32\right)}$$

Take the square root of 7056.

$$x=\frac{-60±84}{2\left(-32\right)}$$

Multiply 2 times -32.

$$x=\frac{-60±84}{-64}$$

Now solve the equation x=\frac{-60±84}{-64} when ± is plus. Add -60 to 84.

$$x=\frac{24}{-64}$$

Reduce the fraction \frac{24}{-64} to lowest terms by extracting and canceling out 8.

$$x=-\frac{3}{8}$$

Now solve the equation x=\frac{-60±84}{-64} when ± is minus. Subtract 84 from -60.

$$x=-\frac{144}{-64}$$

Reduce the fraction \frac{-144}{-64} to lowest terms by extracting and canceling out 16.

$$x=\frac{9}{4}$$

The equation is now solved.

$$x=-\frac{3}{8} x=\frac{9}{4}$$

Steps for Completing the Square

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

$$4\left(x^{2}+6x+9\right)=9\left(2x-1\right)^{2}$$

Use the distributive property to multiply 4 by x^{2}+6x+9.

$$4x^{2}+24x+36=9\left(2x-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-1\right)^{2}.

$$4x^{2}+24x+36=9\left(4x^{2}-4x+1\right)$$

Use the distributive property to multiply 9 by 4x^{2}-4x+1.

$$4x^{2}+24x+36=36x^{2}-36x+9$$

Subtract 36x^{2} from both sides.

$$4x^{2}+24x+36-36x^{2}=-36x+9$$

Combine 4x^{2} and -36x^{2} to get -32x^{2}.

$$-32x^{2}+24x+36=-36x+9$$

Add 36x to both sides.

$$-32x^{2}+24x+36+36x=9$$

Combine 24x and 36x to get 60x.

$$-32x^{2}+60x+36=9$$

Subtract 36 from both sides.

$$-32x^{2}+60x=9-36$$

Subtract 36 from 9 to get -27.

$$-32x^{2}+60x=-27$$

Divide both sides by -32.

$$\frac{-32x^{2}+60x}{-32}=-\frac{27}{-32}$$

Dividing by -32 undoes the multiplication by -32.

$$x^{2}+\frac{60}{-32}x=-\frac{27}{-32}$$

Reduce the fraction \frac{60}{-32} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{15}{8}x=-\frac{27}{-32}$$

Divide -27 by -32.

$$x^{2}-\frac{15}{8}x=\frac{27}{32}$$

Divide -\frac{15}{8}, the coefficient of the x term, by 2 to get -\frac{15}{16}. Then add the square of -\frac{15}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{15}{8}x+\left(-\frac{15}{16}\right)^{2}=\frac{27}{32}+\left(-\frac{15}{16}\right)^{2}$$

Square -\frac{15}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{15}{8}x+\frac{225}{256}=\frac{27}{32}+\frac{225}{256}$$

Add \frac{27}{32} to \frac{225}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{15}{8}x+\frac{225}{256}=\frac{441}{256}$$

Factor x^{2}-\frac{15}{8}x+\frac{225}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{15}{16}\right)^{2}=\frac{441}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{15}{16}\right)^{2}}=\sqrt{\frac{441}{256}}$$

Simplify.

$$x-\frac{15}{16}=\frac{21}{16} x-\frac{15}{16}=-\frac{21}{16}$$

Add \frac{15}{16} to both sides of the equation.

$$x=\frac{9}{4} x=-\frac{3}{8}$$