How do you solve #x^2+12x=-32#?

2 Answers
Apr 1, 2018

Answer:

#x=-4# & #x=-8#

Explanation:

Whenever we're dealing with quadratics, we want to set them equal to zero. We can do this by adding #32# to both sides. Thus, we have:

#x^2+12x+32=0#

Next, we need to think of 2 numbers that sum to #12# and their product is #32#. Since both the sum and product is positive, the two numbers will both be positive.

After some trial and error, we arrive at #4# and #8#, as:

#4+8=12#

and

#4*8=32#

Thus, we have:

#(x+4)(x+8)=0#

The zeroes will be the opposite sign, and we get:

#x=-4# and #x=-8#

Hope this helps!

Apr 1, 2018

Answer:

-4, and - 8

Explanation:

#y = x^2 + 12x + 32 = 0#.
Solve this equation by the new Transforming Method (Socratic, Google Search), Case a = 1.
Find 2 real roots, that are both negative (ac > 0; ab > 0) knowing their sum (- b = - 12) and their product (c = 32).
They are: -4 and - 8.

Note . This method directly gives the 2 real roots. We don't have to factor by grouping and to solve the 2 binomials.