How do you solve #8p ^ { 2} + 38p + 62= - 8p + 6#?

Question

1625 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-22T16:02:24

#p = -1.75 or p = -4#

1) Rearrange so that one side = 0:

#8p^2 + 38p + 62 + 8p - 6 = -8p + 6 + 8p - 6#
(Adding #8p - 6# to both sides)

#8p^2 + 46p + 56 = 0#

#p = (-b +- (sqrt(b^2 - 4ac)))/(2a)#

Where a = 8, b = 46, c = 56.

#p = (-46 +-(sqrt(46^2 - 4 * 8 * 56)))/(2*8)#

#p = (-46 +-sqrt(324))/16#

#p = -1.75 or p = -4#

Answer 2 · 2017-10-22T16:08:42

#p=-4# or #p=(-7)/4#

#8p^2+38p+62=-8p+6#

Subtract #(-8p+6)# from both sides.

#8p^2+38p+62-color(red)((-8p+6))=-8p+6-color(red)((-8p+6))#

#8p^2+46p+56=0#

Divide both sides by #2#

#(8p^2)/color(red)2+(46p)/color(red)2+58/color(red)2=0/color(red)2#

#4p^2+23p+28=0#

Now we have to find two numbers whose product is equal to #28xx4# (#28xx4=112#) and add up to #23#.

The two numbers are #16# and #7#

So #4p^2+23p+28=0# can be written as #4p^2+16p+7p+28=0#

#(4p^2+16p)+(7p+28)=0#

#4p(p+4)+7(p+4)=0#

#(p+4)(4p+7)=0#

So #(p+4)=0# and #(4p+7)=0#

#color(red)1.##p+4=0#

#p=-4#

and

#color(red)2.##4p+7=0#

#4p=-7#

#p=(-7)/4#