# For integers x and y, we know sqrt(x)+ sqrt(y) = 16 and x-y =-128. what is the value of x.

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Expert Answers

thilina-g | Certified Educator

We have two equations

`sqrt(x)+sqrt(y) = 16` --------->1

`x -y = -128` --------->2

we can rewrite 2 as,

`(sqrt(x)+sqrt(y))(sqrt(x)-sqrt(y)) = -128`

16(sqrt(x)-sqrt(y)) =-128

`sqrt(x)-sqrt(y) = -8` we shall call this equation as 3.

Now if we add 1 to 3 you get,

`2sqrt(x) = 16 -8 = 8`

sqrt(x) = 4

x = 16.

then substituting 2, you get,

16-y=-128

y = 16+128 = 144.

Therfore answer is x= 16 and y = 144.

Student Comments

rayasamkiran | Student

sqrt(x) + sqrt(y) = 16

sqrt(x) = 16 - sqrt(y)

Squaring on both sides (a - b)^2 = a^2-2ab+b^2

x = 256 -32*sqrt(y)+y

x - y = 256 - 32*sqrt(y)

But (x - y) is given as -128 -- So substituting the same

-128 = 256 - 32*sqrt(y)

32*sqrt(y) = 384

sqrt(y) = 384/32 = 12

y = 144

x-y = -128 hence x = y - 128 = 144 - 128 = 16

Hence x = 16 and y = 144