What is the square root of w + square root of x = square root of y. ?

Question

What is w, x and y? They are not allowed to be perfect squares or zero.Thanks

Answer 1

let w = w  and  x = w^3, this would result in

     y = w(w+1)^2

*where w is a neither a non-zero nor a perfect sq* - set by qn.

example:

w = 2

x = 2^3

   = 8

y = 2(2+1)^2

   = 2(3)^2

   = 2(9)

   = 18

Arriving at this ans:

Substitute x = w^3 into original eqn,

(√w) + (√w^3) = √y

Factorise √w out,

(√w)[1 + (√w^2)] = √y

Square both sides,

w (1 + w)^2 = y

Ta Da there you go

Love Mathematics (:

Answer 2

w^2+x^2+2 sqrt(wx)=y^2

Answer 3

It does seem as if this question would be easier if it were w^2 + x^2 = y^2 where x,y,w are integer & therefore lead to right angle triangles as another answer suggests.

But if you're sure about the question, then it might start like this;

root(w) + root(x) = root(y)  x,y,w E I

(root(w) + root(x) ^2 = y

w + 2 root(w) root(x) + x = y

so if y is integer 2*root(w)*root(x) must be integer

4*w*x = k^2  k E I   (k is integer)

So let's look at w = 2 & see if there are any solutions;

w = 2

4*2*x = k^2

2*2*2*x = k^2

By inspection, if x = 2 then 2*2*2*x = 2*2 * 2*2 = k*k

Which suggests that root(2) + root(2) = 2*root(2) = root(8)

But also anything that makes 2*2*2*x = k*k will also work.

So 2*2 * 2*2*2*2  ==> root(2) + root(8) = root(2) + 2root(2) = 3root(2) = root(18)

And so on with anything that has 4 as a factor.

I have to go now, but I'm sure you can generalise for any w.

Answer 4

pythagorean triples?