Question

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

Answer 1

`x= -10` AND `x = -11`

Explanation:

`x^2 +21x+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`.

`(10)(11) =110` and `10+11=21`

So, `x^2+21x+110= 0` can be rewritten as:

`(x+10)(x+11)=0`. Now either

`(x+10) =0` and therefore `x=-10`

or `(x+11)=0` and `x=-11`

Answer 2

`x=-10,-11`

Explanation:

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

You can solve this both by factoring and Quadratic formula

`~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`

Factoring

If you have a problem with factoring

Watch this video:

Factor the equation

`rarr(x+10)+(x+12)=0`

Now we can say

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

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

`~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`

This is a Quadratic equation (in form `ax^2+bx+c=0`)

Quadratic formula

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

Where

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

`rarrx=(-21+-sqrt(21^2-4(1)(110)))/(2(1))`

`rarrx=(-21+-sqrt(21^2-4(110)))/(2)`

`rarrx=(-21+-sqrt(441-440))/(2)`

`rarrx=(-21+-sqrt(1))/(2)`

`rarrx=(-21+-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`

`~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`

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

Answer 3

(x + 10)(x + 11)

Explanation:

`y = x^2 + 21x + 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)