Solve 50p-25p^2=9

$$50 p - 25 p ^ { 2 } = 9$$

Solve for p

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

$$p = \frac{9}{5} = 1\frac{4}{5} = 1.8$$

Steps Using Factoring By Grouping

Subtract 9 from both sides.

$$50p-25p^{2}-9=0$$

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

$$-25p^{2}+50p-9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -25p^{2}+ap+bp-9. To find a and b, set up a system to be solved.

$$a+b=50 ab=-25\left(-9\right)=225$$

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 225.

$$1,225 3,75 5,45 9,25 15,15$$

Calculate the sum for each pair.

$$1+225=226 3+75=78 5+45=50 9+25=34 15+15=30$$

The solution is the pair that gives sum 50.

$$a=45 b=5$$

Rewrite -25p^{2}+50p-9 as \left(-25p^{2}+45p\right)+\left(5p-9\right).

$$\left(-25p^{2}+45p\right)+\left(5p-9\right)$$

Factor out -5p in -25p^{2}+45p.

$$-5p\left(5p-9\right)+5p-9$$

Factor out common term 5p-9 by using distributive property.

$$\left(5p-9\right)\left(-5p+1\right)$$

To find equation solutions, solve 5p-9=0 and -5p+1=0.

$$p=\frac{9}{5} p=\frac{1}{5}$$

Steps Using the Quadratic Formula

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.

$$-25p^{2}+50p=9$$

Subtract 9 from both sides of the equation.

$$-25p^{2}+50p-9=9-9$$

Subtracting 9 from itself leaves 0.

$$-25p^{2}+50p-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -25 for a, 50 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-50±\sqrt{50^{2}-4\left(-25\right)\left(-9\right)}}{2\left(-25\right)}$$

Square 50.

$$p=\frac{-50±\sqrt{2500-4\left(-25\right)\left(-9\right)}}{2\left(-25\right)}$$

Multiply -4 times -25.

$$p=\frac{-50±\sqrt{2500+100\left(-9\right)}}{2\left(-25\right)}$$

Multiply 100 times -9.

$$p=\frac{-50±\sqrt{2500-900}}{2\left(-25\right)}$$

Add 2500 to -900.

$$p=\frac{-50±\sqrt{1600}}{2\left(-25\right)}$$

Take the square root of 1600.

$$p=\frac{-50±40}{2\left(-25\right)}$$

Multiply 2 times -25.

$$p=\frac{-50±40}{-50}$$

Now solve the equation p=\frac{-50±40}{-50} when ± is plus. Add -50 to 40.

$$p=-\frac{10}{-50}$$

Reduce the fraction \frac{-10}{-50} to lowest terms by extracting and canceling out 10.

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

Now solve the equation p=\frac{-50±40}{-50} when ± is minus. Subtract 40 from -50.

$$p=-\frac{90}{-50}$$

Reduce the fraction \frac{-90}{-50} to lowest terms by extracting and canceling out 10.

$$p=\frac{9}{5}$$

The equation is now solved.

$$p=\frac{1}{5} p=\frac{9}{5}$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-25p^{2}+50p=9$$

Divide both sides by -25.

$$\frac{-25p^{2}+50p}{-25}=\frac{9}{-25}$$

Dividing by -25 undoes the multiplication by -25.

$$p^{2}+\frac{50}{-25}p=\frac{9}{-25}$$

Divide 50 by -25.

$$p^{2}-2p=\frac{9}{-25}$$

Divide 9 by -25.

$$p^{2}-2p=-\frac{9}{25}$$

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

$$p^{2}-2p+1=-\frac{9}{25}+1$$

Add -\frac{9}{25} to 1.

$$p^{2}-2p+1=\frac{16}{25}$$

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

$$\left(p-1\right)^{2}=\frac{16}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p-1\right)^{2}}=\sqrt{\frac{16}{25}}$$

Simplify.

$$p-1=\frac{4}{5} p-1=-\frac{4}{5}$$

Add 1 to both sides of the equation.

$$p=\frac{9}{5} p=\frac{1}{5}$$