Find the sum $a+b+c+d$.

Consider the following:

$\displaystyle f(x)=x^2-rax-sb$

which has zeros $c,d$.

$\displaystyle g(x)=x^2-rcx-sd$

which has zeros $a,b$.

Given $f(x)\ne g(x)$, find the sum $a+b+c+d$ in terms of $r$ and $s$ only.

Mark's solution:

Consider the following:

$\displaystyle f(x)=x^2-rax-sb$

which has zeros $c,d$.

$\displaystyle g(x)=x^2-rcx-sd$

which has zeros $a,b$.

Given $f(x)\ne g(x)$, find the sum $a+b+c+d$ in terms of $r$ and $s$ only.

We know:

$\displaystyle c+d=ra$

$\displaystyle a+b=rc$

Hence:

$\displaystyle a+b+c+d=r(a+c)$

We also know:

$\displaystyle c^2-sb=a^2-sd$

$\displaystyle sd-sb=a^2-c^2$

$\displaystyle s(d-b)=(a+c)(a-c)$

And we obtain by subtraction of the first 2 equations:

$\displaystyle (d-b)-(a-c)=r(a-c)$

Or:

$\displaystyle d-b=(r+1)(a-c)$

Hence:

$\displaystyle s(r+1)(a-c)=(a+c)(a-c)$

Since $a\ne c$ (otherwise $d=b$ and the two quadratics are identical) there results:

$\displaystyle a+c=s(r+1)$

Hence:

$\displaystyle a+b+c+d=rs(r+1)$

We know:

$\displaystyle c+d=ra$

$\displaystyle a+b=rc$

Hence:

$\displaystyle a+b+c+d=r(a+c)$

We also know:

$\displaystyle c^2-sb=a^2-sd$

$\displaystyle sd-sb=a^2-c^2$

$\displaystyle s(d-b)=(a+c)(a-c)$

And we obtain by subtraction of the first 2 equations:

$\displaystyle (d-b)-(a-c)=r(a-c)$

Or:

$\displaystyle d-b=(r+1)(a-c)$

Hence:

$\displaystyle s(r+1)(a-c)=(a+c)(a-c)$

Since $a\ne c$ (otherwise $d=b$ and the two quadratics are identical) there results:

$\displaystyle a+c=s(r+1)$

Hence:

$\displaystyle a+b+c+d=rs(r+1)$

We know:

$\displaystyle c+d=ra$

$\displaystyle a+b=rc$

Hence:

$\displaystyle a+b+c+d=r(a+c)$

We also know:

$\displaystyle c^2-sb=a^2-sd$

$\displaystyle sd-sb=a^2-c^2$

$\displaystyle s(d-b)=(a+c)(a-c)$

And we obtain by subtraction of the first 2 equations:

$\displaystyle (d-b)-(a-c)=r(a-c)$

Or:

$\displaystyle d-b=(r+1)(a-c)$

Hence:

$\displaystyle s(r+1)(a-c)=(a+c)(a-c)$

Since $a\ne c$ (otherwise $d=b$ and the two quadratics are identical) there results:

$\displaystyle a+c=s(r+1)$

Hence:

$\displaystyle a+b+c+d=rs(r+1)$