How do you solve #x^2 - 8x - 20 = 0# by completing the square?

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Answer 1 · 2016-05-14T09:36:20

#x=10# or #x=-2#

In addition to completing the square we can use the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a=(x-4)# and #b=6# as follows:

#0 = x^2-8x-20#

#=(x-4)^2-4^2-20#

#=(x-4)^2-16-20#

#=(x-4)^2-36#

#=(x-4)^2-6^2#

#=((x-4)-6)((x-4)+6)#

#=(x-10)(x+2)#

So #x=10# or #x=-2#

Answer 2 · 2016-05-14T09:57:53

#x=10, -2#

#color(red)("We know that " (a+b)^2=a^2 +2ab+b^2#

#color(red)("And "(a-b)^2=a^2 -2ab+b^2#

#x^2-8x-20=0#

This can be written as #x^2+(2xx x xx-4)-20=0#

Let #a=x# and #b=-4#

#color(red)("Now, " (x-4)^2=x^2 -8x+16#

#x^2-8x-20=0#

#x^2-8x=20#

Add 16 to both sides:

#x^2-8xcolor(red)(+16)=20color(red)(+16)#

#(x-4)^2=36#

We know that #36=6^2 " and " 36=-6^2 #

Find the square root of both the sides:

#x-4=+-6#

When, #x-4=6#

Add 4 to both sides:

#x-4color(red)(+4)=6color(red)(+4)#

#x=10#

When, #x-4=-6#

Add 4 to both sides:

#x-4color(red)(+4)=-6color(red)(+4)#

#x=-2#