Solve (2x-5)^2=0

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

Solve for x

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Steps Using Factoring By Grouping

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

$$4x^{2}-20x+25=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+25. To find a and b, set up a system to be solved.

$$a+b=-20 ab=4\times 25=100$$

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 100.

$$-1,-100 -2,-50 -4,-25 -5,-20 -10,-10$$

Calculate the sum for each pair.

$$-1-100=-101 -2-50=-52 -4-25=-29 -5-20=-25 -10-10=-20$$

The solution is the pair that gives sum -20.

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

Rewrite 4x^{2}-20x+25 as \left(4x^{2}-10x\right)+\left(-10x+25\right).

$$\left(4x^{2}-10x\right)+\left(-10x+25\right)$$

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

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

Factor out common term 2x-5 by using distributive property.

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

Rewrite as a binomial square.

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

To find equation solution, solve 2x-5=0.

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

Steps Using the Quadratic Formula

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

$$4x^{2}-20x+25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -20 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 4\times 25}}{2\times 4}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 4\times 25}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-20\right)±\sqrt{400-16\times 25}}{2\times 4}$$

Multiply -16 times 25.

$$x=\frac{-\left(-20\right)±\sqrt{400-400}}{2\times 4}$$

Add 400 to -400.

$$x=\frac{-\left(-20\right)±\sqrt{0}}{2\times 4}$$

Take the square root of 0.

$$x=-\frac{-20}{2\times 4}$$

The opposite of -20 is 20.

$$x=\frac{20}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{20}{8}$$

Reduce the fraction \frac{20}{8} to lowest terms by extracting and canceling out 4.

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

Steps for Completing the Square

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

$$4x^{2}-20x+25=0$$

Subtract 25 from both sides. Anything subtracted from zero gives its negation.

$$4x^{2}-20x=-25$$

Divide both sides by 4.

$$\frac{4x^{2}-20x}{4}=-\frac{25}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{20}{4}\right)x=-\frac{25}{4}$$

Divide -20 by 4.

$$x^{2}-5x=-\frac{25}{4}$$

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

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-\frac{25}{4}+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=\frac{-25+25}{4}$$

Add -\frac{25}{4} to \frac{25}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-5x+\frac{25}{4}=0$$

Factor x^{2}-5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-\frac{5}{2}=0 x-\frac{5}{2}=0$$

Add \frac{5}{2} to both sides of the equation.

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

The equation is now solved. Solutions are the same.

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