Question

if x2+y2=0 , x99 + y99=?

Answer 1

We would simplfy the first equation given in the question:

2x+2y=0 this, can be written as:

2 (x+y)=0, I.e. x+y=0…. (1)

Keeping (1) in mind and solvinf the second equation given in the question:

99x+99y , I.e. 99 (x+y)…. (2)

Using (1) in (2):

99 (0)

0…. (3)

From (3) it's clear that:

99x+99y is also equal to 0.

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Aninda Chakrabarty

Answered Mar 11, 2017

x2+y2=0x2+y2=0 implies x and y both are zero(i.e x=0, y=0)

Therefore (x^99)+(y^99) =( 0^99)+(0^99) = 0

Hope that helps

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Amik Raj Behera, studies at DAV Public School, Chandrashekharpur, Bhubaneswar (2019)

Answered Mar 17, 2017

I hope your ‘x2’ means x^2…

So if x^2 + y^2 = 0

and we knoe that square of any real number is positive , it can’t be negative. So let x^2 be A and y^2 be B.

Both A and B are +ve and sum of two +ve nos. can’t be 0.

Therefore we must conclude that both A and B are 0

Now if x^2 and y^2 are zero so x and y must be zero.

0^99 + 0^99 = 0+0= 0

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Nilanjan Mondal, worked at Indian IT Industry

Answered Mar 12, 2017

x^2+y^2=0

Here, sum of two square numbers is 0. A square number(ex- x^2)can't be negative. Hence both x^2=0 and y^2=0

I.e. x=0 and y=0

Hence x^99+y^99

=(0)^99+(0)^99

=0+0

=0

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Abel Thomas Koshy, B.Tech from Christ University Faculty of Engineering (2020)

Answered Jul 3, 2017

If you mean x^2 +y^2=0, then the answer goes like this. The square of any number is always positive and adding 2 positive values never gives you ‘0′. In that case, either the value is x=y=0, which means even x^99 + y^99 = 0, Else, one of the digits must have the complex number i, where i=rootof(-1).

So when you square I, you get a negative value. In that case, x=x(i). That gives you,

x^2 +(xi)^2

= x^2 - x^2

=0.

Anyway, considering the above 2 reasons I have stated, if x^2 + y^2 = 0, then x^99 + y^99 = 0.

Abel

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Saiteja Vemula, Abaqus user

Answered Jul 3, 2017

Continue Reading

**It was not mentioned that the values of x, y are real

since x2+y2=0x2+y2=0

y=ix,−ixy=ix,−ix

consider the 1st root y=ixy=ix

x99+y99=x99+i99x99x99+y99=x99+i99x99

→x99−ix99→x99−ix99

→x99(1−i)→x99(1−i)

considering 2nd root y=−ixy=−ix

x99+y99=x99+(−i)99x99x99+y99=x99+(−i)99x99

→x99+ix99→x99+ix99

→x99(1+i)→x99(1+i)

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Abhishek Pratap, IBO at Visionaries (2017-present)

Answered Jul 5, 2017

It depends on whether we are dealing in complex numbers or real numbers only

Let say, that we are dealing with real numbers then-

x2+y2= 0

Implies that both x & y are real numbers and equal to zero

So,

x99+y99= 0

but when it comes to complex numbers it can have n number of solutions.

I hope this helps you.

Answer 2

Answer:

if x^2+y^2=0 ,fine of the value of x^99+y^99=?