Solve x^2+25x+100=0

$${ x }^{ 2 } +25x+100=0$$

Solve for x

$$x=-20$$

$$x=-5$$

Steps Using Factoring

To solve the equation, factor x^{2}+25x+100 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.

$$a+b=25 ab=100$$

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 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 25.

$$a=5 b=20$$

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

$$\left(x+5\right)\left(x+20\right)$$

To find equation solutions, solve x+5=0 and x+20=0.

$$x=-5 x=-20$$

Steps Using Factoring By Grouping

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+100. To find a and b, set up a system to be solved.

$$a+b=25 ab=1\times 100=100$$

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 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 25.

$$a=5 b=20$$

Rewrite x^{2}+25x+100 as \left(x^{2}+5x\right)+\left(20x+100\right).

$$\left(x^{2}+5x\right)+\left(20x+100\right)$$

Factor out x in the first and 20 in the second group.

$$x\left(x+5\right)+20\left(x+5\right)$$

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

$$\left(x+5\right)\left(x+20\right)$$

To find equation solutions, solve x+5=0 and x+20=0.

$$x=-5 x=-20$$

Steps Using the Quadratic Formula

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x^{2}+25x+100=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 25 for b, and 100 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 25.

$$x=\frac{-25±\sqrt{625-4\times 100}}{2}$$

Multiply -4 times 100.

$$x=\frac{-25±\sqrt{625-400}}{2}$$

Add 625 to -400.

$$x=\frac{-25±\sqrt{225}}{2}$$

Take the square root of 225.

$$x=\frac{-25±15}{2}$$

Now solve the equation x=\frac{-25±15}{2} when ± is plus. Add -25 to 15.

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

Divide -10 by 2.

$$x=-5$$

Now solve the equation x=\frac{-25±15}{2} when ± is minus. Subtract 15 from -25.

$$x=-\frac{40}{2}$$

Divide -40 by 2.

$$x=-20$$

The equation is now solved.

$$x=-5 x=-20$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$x^{2}+25x+100=0$$

Subtract 100 from both sides of the equation.

$$x^{2}+25x+100-100=-100$$

Subtracting 100 from itself leaves 0.

$$x^{2}+25x=-100$$

Divide 25, the coefficient of the x term, by 2 to get \frac{25}{2}. Then add the square of \frac{25}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+25x+\left(\frac{25}{2}\right)^{2}=-100+\left(\frac{25}{2}\right)^{2}$$

Square \frac{25}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+25x+\frac{625}{4}=-100+\frac{625}{4}$$

Add -100 to \frac{625}{4}.

$$x^{2}+25x+\frac{625}{4}=\frac{225}{4}$$

Factor x^{2}+25x+\frac{625}{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}.

$$\left(x+\frac{25}{2}\right)^{2}=\frac{225}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{25}{2}\right)^{2}}=\sqrt{\frac{225}{4}}$$

Simplify.

$$x+\frac{25}{2}=\frac{15}{2} x+\frac{25}{2}=-\frac{15}{2}$$

Subtract \frac{25}{2} from both sides of the equation.

$$x=-5 x=-20$$