If $a,b$ are the roots of $x^2-px+q=0$ and $a',b'$ are the roots of $x^2-p'x+q'=0$, then find the value of $(a-a')^2+(b-b')^2+(a-b')^2+(b-a)^2$

Question

If $a,b$ are the roots of $x^2-px+q=0$ and $a',b'$ are the roots of $x^2-p'x+q'=0$, then find the value of $(a-a')^2+(b-b')^2+(a-b')^2+(b-a)^2$

MY WORK:

I tried using the sum of roots and product of roots rules - then I also tried to expand the expression and use the two sum of roots and product conditions but I couldn't do it. I also tried subtracting and adding the equations but it didn't work.

Answer 1

Your method has a few typos. First off, it should be $a'b'=q'$. Also, at the very last step, the expansion of the square of the second term should read:

$$(b-b')^2=b^2+b'^2\color{red}{-2bb'}$$

@egreg's answer is complete. But, I am leaving my answer open as it points out the mistakes in your attempt.

Answer 2

I don't think you can, unless the last summand is $(b-a')^2$, which leaves the expression unchanged when exchanging $a$ with $b$ and $a'$ with $b'$.

In this case the expansion is $$ a^2-2aa'+a'^2+b^2-2bb'+b'^2+a^2-2ab'+b'^2+b^2-2a'b+a'^2 $$ that can be rewritten as $$ 2(a^2+b^2)+2(a'^2+b'^2)-2a'(a+b)-2b'(a+b) $$ and so $$ 2\bigl((a+b)^2-2ab+(a'+b')^2-2a'b'-(a+b)(a'+b')\bigr) $$