# How do you solve x^2+21x+110=0?

##### 3 Answers
Mar 23, 2016

$x = - 10$ AND $x = - 11$

#### Explanation:

${x}^{2} + 21 x + 110 = 0$ The trinomial in this equation can be factored into the product of two binomials. Look for two numbers which multiply to give a product of 110 and have a sum of 21. In this case the numbers are $10$ and $11$.

$\left(10\right) \left(11\right) = 110$ and $10 + 11 = 21$

So, ${x}^{2} + 21 x + 110 = 0$ can be rewritten as:

$\left(x + 10\right) \left(x + 11\right) = 0$. Now either

$\left(x + 10\right) = 0$ and therefore $x = - 10$

or $\left(x + 11\right) = 0$ and $x = - 11$

Mar 23, 2016

$x = - 10 , - 11$

#### Explanation:

color(blue)(x^2+21x+110=0

You can solve this both by factoring and Quadratic formula

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Factoring

If you have a problem with factoring

Watch this video:

Factor the equation

$\rightarrow \left(x + 10\right) + \left(x + 12\right) = 0$

Now we can say

color(orange)(x+10=0,x+11=0

color(green)(rArrx=-10,-11

$\approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx$

This is a Quadratic equation (in form $a {x}^{2} + b x + c = 0$)

Quadratic formula

color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)

Where

color(red)(a=1,b=21,c=110

$\rightarrow x = \frac{- 21 \pm \sqrt{{21}^{2} - 4 \left(1\right) \left(110\right)}}{2 \left(1\right)}$

$\rightarrow x = \frac{- 21 \pm \sqrt{{21}^{2} - 4 \left(110\right)}}{2}$

$\rightarrow x = \frac{- 21 \pm \sqrt{441 - 440}}{2}$

$\rightarrow x = \frac{- 21 \pm \sqrt{1}}{2}$

$\rightarrow x = \frac{- 21 \pm 1}{2}$

Now we have two solutions

color(indigo)(1))color(indigo)((-21+1)/2=-20/2=-10

color(orange)(2))color(orange)((-21-1)/2=-22/2=-11

$\approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx$

color(blue)( :.ul bar |x=-10,-11|

Mar 23, 2016

(x + 10)(x + 11)

#### Explanation:

$y = {x}^{2} + 21 x + 110 = 0$
Use the new AC Method to factor trinomials (Socratic Search).
Find 2 numbers knowing sum (b = 21) and product (c = 11).
Since ac > 0, they have same sign.
Compose factor pairs of (c = 110) --> ...(5, 22)(10, 11). This sum is 21 = b. Then the numbers are 10 and 11.
Answer: y = (x + 10)(x + 11)