Question

Solve the following quadratic equations by factorization:

`x^2-4sqrt2x+6=0`

Answer 1

We have,

`x^2-4sqrt2x+6=0`

`rArrx^2-3sqrt2x-sqrt2x+6=0`

`rArrx(x-3sqrt2)-sqrt2(x-3sqrt2)=0`

`rArr(x-sqrt2)(x-3sqrt2)=0`

`rArrx-sqrt2=0`

`rArrx=sqrt2`

or

`rArrx-3sqrt2=0`

`rArrx=3sqrt2`

Hence, `x=sqrt2` or `x=3sqrt2`