Solve p^2/(3-2p)^2=4

$$\frac { p ^ { 2 } } { ( 3 - 2 p ) ^ { 2 } } = 4$$

Solve for p

$$p = \frac{6}{5} = 1\frac{1}{5} = 1.2$$

$$p=2$$

Steps Using Factoring By Grouping

Variable p cannot be equal to \frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by \left(2p-3\right)^{2}.

$$p^{2}=4\left(2p-3\right)^{2}$$

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

$$p^{2}=4\left(4p^{2}-12p+9\right)$$

Use the distributive property to multiply 4 by 4p^{2}-12p+9.

$$p^{2}=16p^{2}-48p+36$$

Subtract 16p^{2} from both sides.

$$p^{2}-16p^{2}=-48p+36$$

Combine p^{2} and -16p^{2} to get -15p^{2}.

$$-15p^{2}=-48p+36$$

Add 48p to both sides.

$$-15p^{2}+48p=36$$

Subtract 36 from both sides.

$$-15p^{2}+48p-36=0$$

Divide both sides by 3.

$$-5p^{2}+16p-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -5p^{2}+ap+bp-12. To find a and b, set up a system to be solved.

$$a+b=16 ab=-5\left(-12\right)=60$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 60.

$$1,60 2,30 3,20 4,15 5,12 6,10$$

Calculate the sum for each pair.

$$1+60=61 2+30=32 3+20=23 4+15=19 5+12=17 6+10=16$$

The solution is the pair that gives sum 16.

$$a=10 b=6$$

Rewrite -5p^{2}+16p-12 as \left(-5p^{2}+10p\right)+\left(6p-12\right).

$$\left(-5p^{2}+10p\right)+\left(6p-12\right)$$

Factor out 5p in the first and -6 in the second group.

$$5p\left(-p+2\right)-6\left(-p+2\right)$$

Factor out common term -p+2 by using distributive property.

$$\left(-p+2\right)\left(5p-6\right)$$

To find equation solutions, solve -p+2=0 and 5p-6=0.

$$p=2 p=\frac{6}{5}$$

Steps Using the Quadratic Formula

Variable p cannot be equal to \frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by \left(2p-3\right)^{2}.

$$p^{2}=4\left(2p-3\right)^{2}$$

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

$$p^{2}=4\left(4p^{2}-12p+9\right)$$

Use the distributive property to multiply 4 by 4p^{2}-12p+9.

$$p^{2}=16p^{2}-48p+36$$

Subtract 16p^{2} from both sides.

$$p^{2}-16p^{2}=-48p+36$$

Combine p^{2} and -16p^{2} to get -15p^{2}.

$$-15p^{2}=-48p+36$$

Add 48p to both sides.

$$-15p^{2}+48p=36$$

Subtract 36 from both sides.

$$-15p^{2}+48p-36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -15 for a, 48 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-48±\sqrt{48^{2}-4\left(-15\right)\left(-36\right)}}{2\left(-15\right)}$$

Square 48.

$$p=\frac{-48±\sqrt{2304-4\left(-15\right)\left(-36\right)}}{2\left(-15\right)}$$

Multiply -4 times -15.

$$p=\frac{-48±\sqrt{2304+60\left(-36\right)}}{2\left(-15\right)}$$

Multiply 60 times -36.

$$p=\frac{-48±\sqrt{2304-2160}}{2\left(-15\right)}$$

Add 2304 to -2160.

$$p=\frac{-48±\sqrt{144}}{2\left(-15\right)}$$

Take the square root of 144.

$$p=\frac{-48±12}{2\left(-15\right)}$$

Multiply 2 times -15.

$$p=\frac{-48±12}{-30}$$

Now solve the equation p=\frac{-48±12}{-30} when ± is plus. Add -48 to 12.

$$p=-\frac{36}{-30}$$

Reduce the fraction \frac{-36}{-30} to lowest terms by extracting and canceling out 6.

$$p=\frac{6}{5}$$

Now solve the equation p=\frac{-48±12}{-30} when ± is minus. Subtract 12 from -48.

$$p=-\frac{60}{-30}$$

Divide -60 by -30.

$$p=2$$

The equation is now solved.

$$p=\frac{6}{5} p=2$$

Steps for Completing the Square

Variable p cannot be equal to \frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by \left(2p-3\right)^{2}.

$$p^{2}=4\left(2p-3\right)^{2}$$

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

$$p^{2}=4\left(4p^{2}-12p+9\right)$$

Use the distributive property to multiply 4 by 4p^{2}-12p+9.

$$p^{2}=16p^{2}-48p+36$$

Subtract 16p^{2} from both sides.

$$p^{2}-16p^{2}=-48p+36$$

Combine p^{2} and -16p^{2} to get -15p^{2}.

$$-15p^{2}=-48p+36$$

Add 48p to both sides.

$$-15p^{2}+48p=36$$

Divide both sides by -15.

$$\frac{-15p^{2}+48p}{-15}=\frac{36}{-15}$$

Dividing by -15 undoes the multiplication by -15.

$$p^{2}+\frac{48}{-15}p=\frac{36}{-15}$$

Reduce the fraction \frac{48}{-15} to lowest terms by extracting and canceling out 3.

$$p^{2}-\frac{16}{5}p=\frac{36}{-15}$$

Reduce the fraction \frac{36}{-15} to lowest terms by extracting and canceling out 3.

$$p^{2}-\frac{16}{5}p=-\frac{12}{5}$$

Divide -\frac{16}{5}, the coefficient of the x term, by 2 to get -\frac{8}{5}. Then add the square of -\frac{8}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$p^{2}-\frac{16}{5}p+\left(-\frac{8}{5}\right)^{2}=-\frac{12}{5}+\left(-\frac{8}{5}\right)^{2}$$

Square -\frac{8}{5} by squaring both the numerator and the denominator of the fraction.

$$p^{2}-\frac{16}{5}p+\frac{64}{25}=-\frac{12}{5}+\frac{64}{25}$$

Add -\frac{12}{5} to \frac{64}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$p^{2}-\frac{16}{5}p+\frac{64}{25}=\frac{4}{25}$$

Factor p^{2}-\frac{16}{5}p+\frac{64}{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}.

$$\left(p-\frac{8}{5}\right)^{2}=\frac{4}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p-\frac{8}{5}\right)^{2}}=\sqrt{\frac{4}{25}}$$

Simplify.

$$p-\frac{8}{5}=\frac{2}{5} p-\frac{8}{5}=-\frac{2}{5}$$

Add \frac{8}{5} to both sides of the equation.

$$p=2 p=\frac{6}{5}$$