How can we factorize #8x^2+8sqrt2x+4=0# to solve it?

2 Answers
Dec 4, 2017

#8x^2+8sqrt2x+4=8(x+1/sqrt2)^2=0# and
solution is #+-1/sqrt2#

Explanation:

#8x^2+8sqrt2x+4=0# can be written as

#8(x^2+sqrt2x)+4=0#

or #8(x^2+2xx1/sqrt2x+(1/sqrt2)^2-(1/sqrt2)^2)+4=0#

or #8(x+1/sqrt2)^2-8xx1/2+4=0#

or #8(x+1/sqrt2)^2=0# i.e.

#(x+1/sqrt2)^2=0# or #x+1/sqrt2=0#

As this is an equation, we get #x=-1/sqrt2# as solution.

May 26, 2018

Therefore the factors are:

#(2sqrt2x + 2).(2sqrt2x + 2)#

Value of #x# = #+-1/sqrt2#

Explanation:

Split the middle term so that it adds up to #8sqrt2x# and the same two factors when multiplied gives #8 xx 4 = 32#

Step 1: Find the multiples of 32
#32 = 2 xx 2 xx 2 xx 2 xx 2# = #4sqrt2 xx 4sqrt2#

Now adding #4sqrt2 + 4sqrt2# = #8sqrt2#

So now we have two common factors satisfying the above and these are #4sqrt2# and #4sqrt2#

Step 2: Write #8x^2# in terms of #sqrt2#
#8x^2# = #(2sqrt2) ^2x^2#

Step 3: Re-write the equation

#(2sqrt2)^x^2 +8sqrt2x + 4#

#(2sqrt2)^2x^2 + 4sqrt2x + 4sqrt2x +4#

#2sqrt2x.(2sqrt2x + 2) + 2.(2sqrt2x +2)#

#(2sqrt2x + 2).(2sqrt2x + 2)#

Therefore the factors are:

#(2sqrt2x + 2).(2sqrt2x + 2)#

Value of #x# = #+-1/sqrt2#