10. The equations $x^{2}+a x+b=0$ and $x^{2}+b x+a=0$ both have real roots. It is known that the sum of the squares of the roots of the first equation is equal to the sum of the squares of the roots of the second one, and that $a \neq b$. Find the value of $500+10(a+b)$.
SolUTION
480

Let $p, q$ be the roots of the first equation, and $r, s$ the roots of the second.
Then $(x-p)(x-q)=x^{2}+a x+b$. After expanding the brackets and comparing coefficients we have $a=-p-q$ and $b=p q$.
A similar analysis of the second equation yields $b=-r-s$ and $a=r s$.
Now, $(-p-q)^{2}=p^{2}+q^{2}+2 p q=p^{2}+q^{2}+2 b$.
Similarly, $(-r-s)^{2}=r^{2}+s^{2}+2 r s=r^{2}+s^{2}+2 a$.
Therefore, $(-p-q)^{2}-2 b=(-r-s)^{2}-2 a$ (since $p^{2}+q^{2}=r^{2}+s^{2}$ ).
Therefore, $a^{2}-2 b=b^{2}-2 a$ and hence $a^{2}-b^{2}=2 b-2 a=2(b-a)$.
Therefore $(a-b)(a+b)=-2(a-b)$ and hence $a+b=-2$, since we know that $a-b \neq 0$.
We conclude that $500+10(a+b)=500-20=480$.