Question

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

Answer 1

`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.

Answer 2

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`