Question

Suppose x and y are real numbers and

[x + √(1 + x2)][y + √(1 + y2)] = 1

What is the value of:

(x + y)2​

Answer 1

Step-by-step explanation:

Let us solve for y in terms of x. Suppose:

u = x + √(1 + x2)

Then we can solve for y:

u(y + √(1 + y2)) = 1

y + √(1 + y2) = 1/u

√(1 + y2) = 1/u – y

1 + y2 = (1/u – y)2

1 + y2 = 1/u2 – 2y/u + y2

1 = 1/u2 – 2y/u

2y/u = 1/u2 – 1

y = (1/u – u)/2

Now let’s determine the value of 1/u.

u = x + √(1 + x2)

1/u = 1/(x + √(1 + x2))

We can multiply the numerator and denominator by the conjugate to get:

1/u = (x – √(1 + x2))/[(x + √(1 + x2))(x – √(1 + x2))]

1/u = (x – √(1 + x2))/(x2 – (1 + x2))

1/u = (x – √(1 + x2))/(-1)

1/u = –x + √(1 + x2)

Thus we have:

y = (1/u – u)/2

y = (-x + √(1 + x2) – (x + √(1 + x2))/2

y = (-2x)/2

y = –x

And now we can easily solve the problem:

(x + y)2

= (x – x)2

= 0

I hope you like my answer

Answer 2

the answer is 146 .......