Solve 25*(1-x)(1-x)=16

$$25 \times ( 1 - x ) ( 1 - x ) = 16$$

Solve for x

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

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

Steps Using the Quadratic Formula

Multiply 1-x and 1-x to get \left(1-x\right)^{2}.

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

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

$$25\left(1-2x+x^{2}\right)=16$$

Use the distributive property to multiply 25 by 1-2x+x^{2}.

$$25-50x+25x^{2}=16$$

Subtract 16 from both sides.

$$25-50x+25x^{2}-16=0$$

Subtract 16 from 25 to get 9.

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

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.

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

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}.

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

Square -50.

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

Multiply -4 times 25.

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

Multiply -100 times 9.

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

Add 2500 to -900.

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

Take the square root of 1600.

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

The opposite of -50 is 50.

$$x=\frac{50±40}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{50±40}{50}$$

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

$$x=\frac{90}{50}$$

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

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

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

$$x=\frac{10}{50}$$

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

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

The equation is now solved.

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

Steps for Completing the Square

Multiply 1-x and 1-x to get \left(1-x\right)^{2}.

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

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

$$25\left(1-2x+x^{2}\right)=16$$

Use the distributive property to multiply 25 by 1-2x+x^{2}.

$$25-50x+25x^{2}=16$$

Subtract 25 from both sides.

$$-50x+25x^{2}=16-25$$

Subtract 25 from 16 to get -9.

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

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.

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

Divide both sides by 25.

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

Dividing by 25 undoes the multiplication by 25.

$$x^{2}+\left(-\frac{50}{25}\right)x=-\frac{9}{25}$$

Divide -50 by 25.

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

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.

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

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

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

Factor x^{2}-2x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 1 to both sides of the equation.

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