How do you solve #a^2-2a-224=0#?

2 Answers
Jul 8, 2015

Answer:

Solve #y = x^2 - 2x - 224 = 0#

Explanation:

#y = x^2 - 2x - 224 = 0#.
I use the new Transforming Method (Google Yahoo Search)
Find 2 numbers knowing sum (2) and product (-224). Roots have different signs.
Factor pairs of (-224) -> ...(-8, 28)(-14, 16). This sum is 2 = -b. Then, the 2 real roots are: #- 14 and 16.#

Jul 8, 2015

Answer:

Factor the trinomial as the product of two binomials. Set each binomial equal to zero, and solve for #a#.

Explanation:

#a^2-2a-224=0#

Find two numbers that when added equal #-2#, and when multiplied equal #-224#. The numbers #-16# and #14# fit the pattern.

#a^2-2a-224=0# =

#(a-16)(a+14)=0#

Set #(a-16)# equal to zero and solve for #a#.

#a-16=0#

#a=16

Set #(a+14)# equal to zero and solve for #a#.

#a+14=0#

#a=-14#

#a=16, -14#