Solve 25(x-1)^2-9=0

$$25 ( x - 1 ) ^ { 2 } - 9 = 0$$

Solve for x

$$x = \frac{8}{5} = 1\frac{3}{5} = 1.6$$

$$x=\frac{2}{5}=0.4$$

Steps Using Factoring By Grouping

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

$$25\left(x^{2}-2x+1\right)-9=0$$

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

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

Subtract 9 from 25 to get 16.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 25x^{2}+ax+bx+16. To find a and b, set up a system to be solved.

$$a+b=-50 ab=25\times 16=400$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 400.

$$-1,-400 -2,-200 -4,-100 -5,-80 -8,-50 -10,-40 -16,-25 -20,-20$$

Calculate the sum for each pair.

$$-1-400=-401 -2-200=-202 -4-100=-104 -5-80=-85 -8-50=-58 -10-40=-50 -16-25=-41 -20-20=-40$$

The solution is the pair that gives sum -50.

$$a=-40 b=-10$$

Rewrite 25x^{2}-50x+16 as \left(25x^{2}-40x\right)+\left(-10x+16\right).

$$\left(25x^{2}-40x\right)+\left(-10x+16\right)$$

Factor out 5x in the first and -2 in the second group.

$$5x\left(5x-8\right)-2\left(5x-8\right)$$

Factor out common term 5x-8 by using distributive property.

$$\left(5x-8\right)\left(5x-2\right)$$

To find equation solutions, solve 5x-8=0 and 5x-2=0.

$$x=\frac{8}{5} x=\frac{2}{5}$$

Steps Using the Quadratic Formula

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

$$25\left(x^{2}-2x+1\right)-9=0$$

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

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

Subtract 9 from 25 to get 16.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, -50 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-50\right)±\sqrt{\left(-50\right)^{2}-4\times 25\times 16}}{2\times 25}$$

Square -50.

$$x=\frac{-\left(-50\right)±\sqrt{2500-4\times 25\times 16}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-\left(-50\right)±\sqrt{2500-100\times 16}}{2\times 25}$$

Multiply -100 times 16.

$$x=\frac{-\left(-50\right)±\sqrt{2500-1600}}{2\times 25}$$

Add 2500 to -1600.

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

Take the square root of 900.

$$x=\frac{-\left(-50\right)±30}{2\times 25}$$

The opposite of -50 is 50.

$$x=\frac{50±30}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{50±30}{50}$$

Now solve the equation x=\frac{50±30}{50} when ± is plus. Add 50 to 30.

$$x=\frac{80}{50}$$

Reduce the fraction \frac{80}{50} to lowest terms by extracting and canceling out 10.

$$x=\frac{8}{5}$$

Now solve the equation x=\frac{50±30}{50} when ± is minus. Subtract 30 from 50.

$$x=\frac{20}{50}$$

Reduce the fraction \frac{20}{50} to lowest terms by extracting and canceling out 10.

$$x=\frac{2}{5}$$

The equation is now solved.

$$x=\frac{8}{5} x=\frac{2}{5}$$

Steps for Completing the Square

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

$$25\left(x^{2}-2x+1\right)-9=0$$

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

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

Subtract 9 from 25 to get 16.

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

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 25.

$$\frac{25x^{2}-50x}{25}=-\frac{16}{25}$$

Dividing by 25 undoes the multiplication by 25.

$$x^{2}+\left(-\frac{50}{25}\right)x=-\frac{16}{25}$$

Divide -50 by 25.

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

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.

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

Add -\frac{16}{25} to 1.

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

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}.

$$\left(x-1\right)^{2}=\frac{9}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{\frac{9}{25}}$$

Simplify.

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

Add 1 to both sides of the equation.

$$x=\frac{8}{5} x=\frac{2}{5}$$