Solve 16*p^2=(1-p)^2

$$16 \cdot p ^ { 2 } = ( 1 - p ) ^ { 2 }$$

Solve for p

$$p=-\frac{1}{3}\approx -0.333333333$$

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

Steps Using Factoring By Grouping

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

$$16p^{2}=1-2p+p^{2}$$

Subtract 1 from both sides.

$$16p^{2}-1=-2p+p^{2}$$

Add 2p to both sides.

$$16p^{2}-1+2p=p^{2}$$

Subtract p^{2} from both sides.

$$16p^{2}-1+2p-p^{2}=0$$

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

$$15p^{2}-1+2p=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$15p^{2}+2p-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 15p^{2}+ap+bp-1. To find a and b, set up a system to be solved.

$$a+b=2 ab=15\left(-1\right)=-15$$

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 -15.

$$-1,15 -3,5$$

Calculate the sum for each pair.

$$-1+15=14 -3+5=2$$

The solution is the pair that gives sum 2.

$$a=-3 b=5$$

Rewrite 15p^{2}+2p-1 as \left(15p^{2}-3p\right)+\left(5p-1\right).

$$\left(15p^{2}-3p\right)+\left(5p-1\right)$$

Factor out 3p in 15p^{2}-3p.

$$3p\left(5p-1\right)+5p-1$$

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

$$\left(5p-1\right)\left(3p+1\right)$$

To find equation solutions, solve 5p-1=0 and 3p+1=0.

$$p=\frac{1}{5} p=-\frac{1}{3}$$

Steps Using the Quadratic Formula

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

$$16p^{2}=1-2p+p^{2}$$

Subtract 1 from both sides.

$$16p^{2}-1=-2p+p^{2}$$

Add 2p to both sides.

$$16p^{2}-1+2p=p^{2}$$

Subtract p^{2} from both sides.

$$16p^{2}-1+2p-p^{2}=0$$

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

$$15p^{2}-1+2p=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$15p^{2}+2p-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 15 for a, 2 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-2±\sqrt{2^{2}-4\times 15\left(-1\right)}}{2\times 15}$$

Square 2.

$$p=\frac{-2±\sqrt{4-4\times 15\left(-1\right)}}{2\times 15}$$

Multiply -4 times 15.

$$p=\frac{-2±\sqrt{4-60\left(-1\right)}}{2\times 15}$$

Multiply -60 times -1.

$$p=\frac{-2±\sqrt{4+60}}{2\times 15}$$

Add 4 to 60.

$$p=\frac{-2±\sqrt{64}}{2\times 15}$$

Take the square root of 64.

$$p=\frac{-2±8}{2\times 15}$$

Multiply 2 times 15.

$$p=\frac{-2±8}{30}$$

Now solve the equation p=\frac{-2±8}{30} when ± is plus. Add -2 to 8.

$$p=\frac{6}{30}$$

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

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

Now solve the equation p=\frac{-2±8}{30} when ± is minus. Subtract 8 from -2.

$$p=-\frac{10}{30}$$

Reduce the fraction \frac{-10}{30} to lowest terms by extracting and canceling out 10.

$$p=-\frac{1}{3}$$

The equation is now solved.

$$p=\frac{1}{5} p=-\frac{1}{3}$$

Steps for Completing the Square

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

$$16p^{2}=1-2p+p^{2}$$

Add 2p to both sides.

$$16p^{2}+2p=1+p^{2}$$

Subtract p^{2} from both sides.

$$16p^{2}+2p-p^{2}=1$$

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

$$15p^{2}+2p=1$$

Divide both sides by 15.

$$\frac{15p^{2}+2p}{15}=\frac{1}{15}$$

Dividing by 15 undoes the multiplication by 15.

$$p^{2}+\frac{2}{15}p=\frac{1}{15}$$

Divide \frac{2}{15}, the coefficient of the x term, by 2 to get \frac{1}{15}. Then add the square of \frac{1}{15} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$p^{2}+\frac{2}{15}p+\left(\frac{1}{15}\right)^{2}=\frac{1}{15}+\left(\frac{1}{15}\right)^{2}$$

Square \frac{1}{15} by squaring both the numerator and the denominator of the fraction.

$$p^{2}+\frac{2}{15}p+\frac{1}{225}=\frac{1}{15}+\frac{1}{225}$$

Add \frac{1}{15} to \frac{1}{225} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$p^{2}+\frac{2}{15}p+\frac{1}{225}=\frac{16}{225}$$

Factor p^{2}+\frac{2}{15}p+\frac{1}{225}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(p+\frac{1}{15}\right)^{2}=\frac{16}{225}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p+\frac{1}{15}\right)^{2}}=\sqrt{\frac{16}{225}}$$

Simplify.

$$p+\frac{1}{15}=\frac{4}{15} p+\frac{1}{15}=-\frac{4}{15}$$

Subtract \frac{1}{15} from both sides of the equation.

$$p=\frac{1}{5} p=-\frac{1}{3}$$