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

3 Answers
Mar 23, 2016

Answer:

#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#

Mar 23, 2016

Answer:

#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

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

Now we can say

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

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

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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|#

Mar 23, 2016

Answer:

(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)