Math, asked by fathimashana7777, 1 year ago

Whole root of (√3−1)^(2)−(√3−1)^(2)+4^(2)

Answers

Answered by Anonymous
0

Answer:

ans 4

Step-by-step explanation:

√(√3-1)^2-(√3-1)^2+(4)^2

√ (4)^2 =4

hope it's helpful.

Answered by Ritiksuglan
0

hey mates your answer is here

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

=

a

2

+

2

a

b

2

+

2

b

2

So

a

2

+

2

b

2

=

3

and

2

a

b

=

2

a

b

=

1

These are two simultaneous equations for

a

and

b

.

Rearrange the second:

b

=

1

a

Substitute into the first:

a

2

+

2

a

2

=

3

a

4

+

2

=

3

a

2

a

4

3

a

2

+

2

=

0

Note that this is a quadratic in

a

2

:

(

a

2

)

2

3

(

a

2

)

+

2

=

0

Factorise:

(

a

2

2

)

(

a

2

1

)

=

0

This gives us two possible solutions for

a

2

:

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

=

1

a

b

=

¯¯¯¯¯

+

1

2

=

¯¯¯¯¯

+

1

2

2

and

¯¯¯¯

+

1

.

So we have the four solution pairs

(

a

,

b

)

:

(

2

,

1

2

2

)

(

2

,

1

2

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

1

2

2

2

=

1

+

2

2

+

1

2

2

2

=

1

2

1

2

1

+

2

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

may be it's helpful for you

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