Question

???
$\sqrt{2 - \sqrt{3 } } = ?$ Please help​

Answer 1

Answer:

hope it's help you

plz mark as brain list

Answer 2

Answer:

Check your ans

Step-by-step explanation:

The answer above is completely correct, but it may help the student to see how to reach it:

We seek a square root of 3−2√2. This must have the form a+b√2.

(a+b√2)2=a2+2ab√2+2b2

So

a2+2b2=3

and

2ab=−2⇒ab=−1

These are two simultaneous equations for a and b.

Rearrange the second:

b=−1a

Substitute into the first:

a2+2a2=3

a4+2=3a2

a4−3a2+2=0

Note that this is a quadratic in a2:

(a2)2−3(a2)+2=0

Factorise:

(a2−2)(a2−1)=0

This gives us two possible solutions for a2: 2 and 1, and so the four solutions for a: ±√2 and ±1.

We are looking for integer solutions for a, and so ±1 are possible solutions. But the other two are possible too - they can simply be folded in to the √2 term. This wouldn't have been possible if we'd had the root of some other number in the solution for a, but this solution is a special case.

Now use the second equation to deduce the four equivalent solutions for b:

b=−1a

b=¯¯¯¯¯+1√2=¯¯¯¯¯+12√2 and ¯¯¯¯+1.

So we have the four solution pairs (a,b):

(√2,−12√2)

(−√2,12√2)

(1,−1)

(−1,1)

This is a bit suspicious - we expect only two solutions, positive and negative square roots, so we wonder if some of these are identical to each other: When we substitute them in to our desired expression a+b√2, we get:

√2−12√2√2=−1+√2

−√2+12√2√2=1−√

So the two solutions with √2 are identical to the two simpler solutions, so we can get rid of them. We now have two solutions, positive and negative square roots:

1−√2

−1+√2

When we take the written square root of a quantity, it is implied that the desired root is the positive root, the "principal value" of the square root function. So we take the single solution that comes from a=+1:

1−√2

Double check: Make sure that this produces the desired answer:

(1−√2)2=1−2√2+2=3−2√2

Hope it's helpful

MARK ME AS BRAINLIEST