Solve x^2=-10x-25

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

Solve for x

$$x=-5$$

Steps Using Factoring

Add 10x to both sides.

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

Add 25 to both sides.

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

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 positive, a and b are both positive. 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

Add 10x to both sides.

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

Add 25 to both sides.

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

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 positive, a and b are both positive. 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

Add 10x to both sides.

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

Add 25 to both sides.

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

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}.

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

Square 10.

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

Multiply -4 times 25.

$$x=\frac{-10±\sqrt{100-100}}{2}$$

Add 100 to -100.

$$x=\frac{-10±\sqrt{0}}{2}$$

Take the square root of 0.

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

Divide -10 by 2.

$$x=-5$$

Steps for Completing the Square

Add 10x to both sides.

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

Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+10x+5^{2}=-25+5^{2}$$

Square 5.

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

Add -25 to 25.

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

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.

$$x+5=0 x+5=0$$

Subtract 5 from both sides of the equation.

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

The equation is now solved. Solutions are the same.

$$x=-5$$