Sum of two numbers is $32$ and sum of their squares is $904$. What are the numbers?

Question

According to the problem I've tried

$$x+y=32.........(1)$$

$$x^2+y^2=904........(2)$$

Now, I know $x^2+y^2$ can be written as- $$(x+y)^2+2xy$$

But don't know how to solve further steps. Please help.

Answer 1

$$y=32-x$$

$$x^2+(32-x)^2=904$$

$$2x^2-64x+32^2-904=0$$

$$2x^2-64x+120=0$$

$$x^2-32x+60=0$$

$$(x-2)(x-30)=0$$

Are you able to solve the remaining?

Answer 2

We can write $x^2+y^2$ as $(x+y)^2-2xy$. We know that $x+y=32$. So, we thus get, $2xy =120 \Rightarrow xy=60$.

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Now use the fact that $(x-y)^2 = (x+y)^2-4xy$. Solve for $(x-y)$ and thus we can get the answer for $x$ by adding $(x-y)$ and $(x+y)$. Hope it helps.

Answer 3

$x+y=32.........[1]$

$x^2+y^2=904$

$\implies (x+y)^2–2xy=904$

$\implies 2xy=(x+y)^2–904$

$\implies 2xy=32^2–904$

$\implies xy=60$

$(x-y)^2=(x+y)^2–4xy$

$\implies (x-y)^2=32^2–4\times 60$

$\implies x-y=28.......[2]$

[I’m not writing $\pm 28$, since I am considering $x>y$]

Now, solving [1] and [2] simultaneously gives

$2x=60$

$\implies x=30$

$\implies y=2$