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$$x^2 + xy + y^2 = a$$

$$x + y = b$$

$xy = ?$

I tried this and did this:

$xy = a - x^2 - y^2$

$xy = a - (x^2 + y^2)$

$xy = a - (x + y)(x - y)$

$xy = a - b(x - y)$

At this point I can't think of anything to do to represent the $x - y$ part in terms of $a$ and $b$.

Any help would be appreciated.

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5 Answers

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$b^2=(x+y)^2=x^2+2xy+y^2=a+xy$. Therefore $xy=b^2-a$.

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We are given \begin{align}x^2+xy+y^2&=a\tag{1}\\ x+y&=b\tag{2}\end{align}

We can square the equation $(2)$ to give us \begin{align}b^2&=(x+y)^2\\ &=x^2+2xy+y^2\end{align}

Therefore, we now have \begin{align}a&=x^2+xy+y^2\tag{1}\\ b^2&=x^2+2xy+y^2\tag{3}\end{align}

We can rearrange equation $(3)$ and substitute equation $(1)$ into it to give us the answer: \begin{align}b^2&=(x^2+xy+y^2)+xy\\ b^2&=a+xy\\ xy&=b^2-a\end{align}

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$x^2 + xy + y^2 = a$ ....(1)

$x + y = b$

Squaring both sides,

$x^2 + y^2 + 2xy = b^2$

$x^2 + y^2 = b^2 - 2xy$

Put value of $x^2 + y^2$ in equation (1),

$b^2 - 2xy + xy = a$

$b^2 - xy = a$

$xy = b^2 - a$

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plugging nyour second equation $$y=b-x$$ in the first equation we get $$x^2+x(b-x)+(b-x)^2=a$$ and it follows $$x^2-bx+b^2-a=0$$ solve this quadratic equation and check the answers

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I like this trick: $$x^2+xy+y^2=\frac{3(x+y)^2+(x-y)^2}{4},$$ so $$(x-y)^2=4a-3b^2,$$ $$x-y=\pm\sqrt{4a-3b^2}$$ etc.

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