How do you find the roots for #f(x) = x^2 + 12x + 20#?

3 Answers
Apr 1, 2018

Answer:

#x_1=-2 or x_2=-10#

Explanation:

#f(x) = x^2 + 12x + 20#

Let #f(x)=0#
#0 = x^2 + 12x + 20#
#0 = x^2 + 2*6*x + 6^2-6^2+20#
#0 = (x+6)^2-36+20= (x+6)^2-16|+16#
#16 = (x+6)^2|sqrt()#
#+-sqrt(16)=x+6|-6#
#-6+-4=x_(1,2)#
#x_1=-2 or x_2=-10#

Apr 1, 2018

Answer:

  • #x = -10#
  • #x = -2#

Explanation:

#f(x) = x^2 +12x +20#

The roots are the x-intercepts. These occur where #f(x) = 0#.

#0 = x^2 + 12x + 20#

Now we can factor the right side. We need to find factors of #20# that when summed give us #12#.

Factors of #20#: #(1,20), (2,10),(4,5)#

If we sum each pair, the one that gives us #12# is #(2,10)#.

Hence, we factor as:

#0 = (x+10)(x+2)#

So the roots are found as:

  • #x+10=0 -> x = -10#
  • #x+2=0 -> x = -2#

Hence:

  • #x = -10#
  • #x = -2#
Apr 1, 2018

Answer:

-2 and - 10

Explanation:

#f(x) = x^2 + 12x + 20 = 0#.
Find 2 real roots, that are both negative (ac > 0; ab > 0), knowing their sum (- b = - 12) and their product (c = 20).
They are: -2 and - 10.

Note . When a = 1, we don't have to do factoring by grouping and solving the 2 binomials.