Question

Solve the following quadratic equations by factorization:

`sqrt2x^2-3x-2sqrt2=0`

Answer 1

We have been given

`sqrt2x^2-3x-2sqrt2=0`

`sqrt2x^2-4x+x-2sqrt2=0`

`sqrt2x(x-2sqrt2)+1(x-2sqrt2)=0`

`(x-2sqrt2)(sqrt2x+1)=0`

Therefore,

`x-2sqrt2=0`

`x=2sqrt2`

or,

`sqrt2x+1=0`

`sqrt2x=-1`

`x=(-1)/sqrt2`

Hence, `x=2sqrt2` or `x=(-1)/sqrt2`