How do you solve #g^2+4g-32=0#?

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Answer 1 · 2016-10-31T14:43:07

# x = ^+4 #

# x = - 8#

#Ax^2 +Bx +C = g^2 +4g -32#

Because the C value of -32 is negative, it must have a positive and a negative factor.

The B value of + 4 shows that the positive factor must be four larger than the negative factor.

Possible factors are: #1 xx 32," "2 xx 16," " 4 xx 8#

Only #4 xx 8# has a difference of #4#, so #4 and 8# are the factors to use to solve this equation.

The #8# must be positive and the #4# negative. So the equation is

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

Each factor could be equal to 0, so solve for each binomial

# x- 4 = 0" "# add four to both sides

# x- 4 +4 = 0 + 4 # resulting in

#x = + 4 #

# x + 8 = 0" "# subtract 8 from both sides

# x + 8 -8 = 0 -8 #

#x = -8#