Solve (5x-1)^2=0

$$( 5 x - 1 ) ^ { 2 } = 0$$

Solve for x

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

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-1\right)^{2}.

$$25x^{2}-10x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 25x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-10 ab=25\times 1=25$$

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 25.

$$-1,-25 -5,-5$$

Calculate the sum for each pair.

$$-1-25=-26 -5-5=-10$$

The solution is the pair that gives sum -10.

$$a=-5 b=-5$$

Rewrite 25x^{2}-10x+1 as \left(25x^{2}-5x\right)+\left(-5x+1\right).

$$\left(25x^{2}-5x\right)+\left(-5x+1\right)$$

Factor out 5x in the first and -1 in the second group.

$$5x\left(5x-1\right)-\left(5x-1\right)$$

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

$$\left(5x-1\right)\left(5x-1\right)$$

Rewrite as a binomial square.

$$\left(5x-1\right)^{2}$$

To find equation solution, solve 5x-1=0.

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

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-1\right)^{2}.

$$25x^{2}-10x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, -10 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 25}}{2\times 25}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 25}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-\left(-10\right)±\sqrt{100-100}}{2\times 25}$$

Add 100 to -100.

$$x=\frac{-\left(-10\right)±\sqrt{0}}{2\times 25}$$

Take the square root of 0.

$$x=-\frac{-10}{2\times 25}$$

The opposite of -10 is 10.

$$x=\frac{10}{2\times 25}$$

Multiply 2 times 25.

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

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

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

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-1\right)^{2}.

$$25x^{2}-10x+1=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$25x^{2}-10x=-1$$

Divide both sides by 25.

$$\frac{25x^{2}-10x}{25}=-\frac{1}{25}$$

Dividing by 25 undoes the multiplication by 25.

$$x^{2}+\left(-\frac{10}{25}\right)x=-\frac{1}{25}$$

Reduce the fraction \frac{-10}{25} to lowest terms by extracting and canceling out 5.

$$x^{2}-\frac{2}{5}x=-\frac{1}{25}$$

Divide -\frac{2}{5}, the coefficient of the x term, by 2 to get -\frac{1}{5}. Then add the square of -\frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{5}x+\left(-\frac{1}{5}\right)^{2}=-\frac{1}{25}+\left(-\frac{1}{5}\right)^{2}$$

Square -\frac{1}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=\frac{-1+1}{25}$$

Add -\frac{1}{25} to \frac{1}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=0$$

Factor x^{2}-\frac{2}{5}x+\frac{1}{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}.

$$\left(x-\frac{1}{5}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{5}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-\frac{1}{5}=0 x-\frac{1}{5}=0$$

Add \frac{1}{5} to both sides of the equation.

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

The equation is now solved. Solutions are the same.

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