Problem Solving

Expand: sqrt2(3+sqrt3)

Answer is 3sqrt2 + sqrt6

My question is, why don't we distribute the sqrt 2 across both terms --> 3sqrt2 + sqrt 6. Then break sqrt 6 into (sqrt 2)(sqrt 3). From here we would take 1sqrt 2 and add it to our 3sqrt2 to get 4sqrt2+sqrt3.

Am I missing something?

Is it because (sqrt2)(sqrt3) is considered a single term and cannot be split up? i.e. cannot take 1sqrt2 from the sqrt3 and add it to the 3sqrt2.?

(sqrt2)[3+(sqrt3)]

= 3*sqrt(2) + sqrt(2)*sqrt(3)

= 3*sqrt(2) + sqrt(6)

Steven your doubt is something like: 2*3 = 6 = 2*3

At first you are combining them and then you are splitting them.

3sqrt2 + sqrt 6 --> here if you again take break sqrt(6) into sqrt(2) & sqrt(3).. and then take common.. then would will again reach to starting point; that is, you are reversing the steps you performed


(sqrt2)[3+(sqrt3)]

= 3*sqrt(2) + sqrt(2)*sqrt(3)

= 3*sqrt(2) + sqrt(6)

= 3*sqrt(2) + sqrt(2)*sqrt(3)

(sqrt2)[3+(sqrt3)]

stevennu wrote:Expand: sqrt2(3+sqrt3)

Answer is 3sqrt2 + sqrt6

My question is, why don't we distribute the sqrt 2 across both terms --> 3sqrt2 + sqrt 6. Then break sqrt 6 into (sqrt 2)(sqrt 3). From here we would take 1sqrt 2 and add it to our 3sqrt2 to get 4sqrt2+sqrt3.

Am I missing something?

I hope it's (3 + √3) √2, then 3√2 + √2√3 is the next step. Now, how on earth can we snatch that √2 from √3 to nurture 3√2?

You are missing this point...

We can pick and add only what's free for it. Our √2 is already engaged with √3, and the name of that nuptial knot is "multiplied together", means "it's not free for it".

3√2 + √6 is the correct answer in its correct form.

Thanks all! I really appreciate the thorough responses. I'd buy you all a drink if I could.

stevennu wrote:Thanks all! I really appreciate the thorough responses. I'd buy you all a drink if I could.

Vodka or Country Made?