Solve 0=x^2-10x+25

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

Solve for x

$$x=5$$

Steps Using Factoring

Swap sides so that all variable terms are on the left hand side.

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

To solve the equation, factor x^{2}-10x+25 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-10 ab=25$$

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 factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

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

Rewrite as a binomial square.

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

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

$$x=5$$

Steps Using Factoring By Grouping

Swap sides so that all variable terms are on the left hand side.

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

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

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

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

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

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

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

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

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

Rewrite as a binomial square.

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

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

$$x=5$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

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

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

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

Square -10.

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

Multiply -4 times 25.

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

Add 100 to -100.

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

Take the square root of 0.

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

The opposite of -10 is 10.

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

Divide 10 by 2.

$$x=5$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

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

Factor x^{2}-10x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-5=0 x-5=0$$

Add 5 to both sides of the equation.

$$x=5 x=5$$

The equation is now solved. Solutions are the same.

$$x=5$$