How do you solve #8x^{2} + 21= - 59x#?

2 Answers
Mar 14, 2018

#x = -7, -0.375#

Explanation:

#8x^2 + 21 = -59 x#

add #59x# to both sides

#8x^2 + 59x + 21#

factor by grouping

#color(white)(.11) + 59#
#color(white)(.11) xx 168#
. . . . . . . . . .
#color(white)(.) color(white)(1) 1 xx 168# #=> 1+168 = 169#
#color(white)(.)color(white)(1) 2 xx 84 color(white)(1) # #=> 2 + 84 = 86#
#color(white)(.)color(white)(1) 3 xx 56 color(white)(1) # #=> 3 + 56 = color(red)(59)#

#(8x^2 + 3x) + (56x + 21)#
#x(8x+3) + 7(8x + 3)#
#(x+7)(8x+3)#
Set each component to #0# and solve for #x#:

#* * * * * * * * * * * * * * * * * * * * * * * * * #

Case 1

#8x + 3 = 0#

#8x = -3#

#x = -3/8#

#* * * * * * * * * * * * * * * * * * * * * * * * * #

Case 2

#x + 7 = 0#

#x = -7#

So #x = -7# and #-3/8# or #-0.375#

Let's graph the equation and see if we are right

graph{y = 8x^2 + 59x + 21}

We are! Nice job

Mar 14, 2018

#-3/8, -7#

Explanation:

By factoring:
The factors of 8 are (8,1) or (4,2)
The factors of 21 are (21,1) or (7,3)

#21*8=168# Too large
#21*4=84# Too large
#7*8=56# Worth considering

Rearrange formula to equal zero
#8x^2+59x+21=0#
Since all signs are +, all factors will be +

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

By trial-and-error:
#(8x+3)(x+7)=8x^2+56x+3x+21=8x^2+59x+21#

So, #8x+3=0# and #x+7=0# Solve each for #x#

#8x+3=0#
#8x=-3#
#x=-3/8#

And

#x+7=0#
#x=-7#