How do you solve #1/2(x+8)^2=14#?

2 Answers
Jan 28, 2017

Answer:

#x = 2sqrt(7) - 8#

Explanation:

#1/"2"(x + 8)^2 = 14#

Multiplying by 2 on both sides

#cancel(2) × 1/cancel("2")(x + 8)^2 = 2 × 14#

#(x + 8)^2 = 28#

Square root both side

#sqrt((x + 8)^2# = #sqrt(28)#

#x + 8 = +-2sqrt(7)#

#x = 2sqrt(7) - 8 and x =- 2sqrt(7) - 8#

Answer:

#x=-8+-2sqrt7#

Explanation:

#1/2 (x+8)^2=14#

I want to expand the squared term, so to do that I'll first multiply both sides by 2:

#1/2 color(red)(xx2) (x+8)^2=14color(red)(xx2)#

#(x+8)^2=28#

#x^2+16x+64=28#

#x^2+16x+64color(red)(-28)=28color(red)(-28)#

#x^2+16x+36=0#

I don't see any easy factors, so let's use the quadratic formula. The general formula is:

#x=(-color(green)b+-sqrt(color(green)b^2-4color(blue)acolor(brown)c))/(2color(blue)a)#

and relates to a trinomial this way:

#color(blue)ax^2+color(green)bx+color(brown)c#

#x=(-16+-sqrt(16^2-4(1)(36)))/(2(1))#

#x=(-16+-sqrt(256-144))/2#

#x=(-16+-sqrt(112))/2=(-16+-sqrt(16xx7))/2=(-16+-4sqrt(7))/2=-8+-2sqrt7#

And we can see these two solutions in the graphing of both sides of the original equation:

graph{(y-(1/2)(x+8)^2)(y-0x-14)=0 [-16.69, 3.31, 9.14, 19.136]}