[MATH]f(x)=x^2-rax-sb[/MATH]
which has zeros \(c,d\).
[MATH]g(x)=x^2-rcx-sd[/MATH]
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:
[MATH]f(x)=x^2-rax-sb[/MATH]
which has zeros \(c,d\).
[MATH]g(x)=x^2-rcx-sd[/MATH]
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:
[MATH]c+d=ra[/MATH]
[MATH]a+b=rc[/MATH]
Hence:
[MATH]a+b+c+d=r(a+c)[/MATH]
We also know:
[MATH]c^2-sb=a^2-sd[/MATH]
[MATH]sd-sb=a^2-c^2[/MATH]
[MATH]s(d-b)=(a+c)(a-c)[/MATH]
And we obtain by subtraction of the first 2 equations:
[MATH](d-b)-(a-c)=r(a-c)[/MATH]
Or:
[MATH]d-b=(r+1)(a-c)[/MATH]
Hence:
[MATH]s(r+1)(a-c)=(a+c)(a-c)[/MATH]
Since \(a\ne c\) (otherwise \(d=b\) and the two quadratics are identical) there results:
[MATH]a+c=s(r+1)[/MATH]
Hence:
[MATH]a+b+c+d=rs(r+1)[/MATH]
We know:
[MATH]c+d=ra[/MATH]
[MATH]a+b=rc[/MATH]
Hence:
[MATH]a+b+c+d=r(a+c)[/MATH]
We also know:
[MATH]c^2-sb=a^2-sd[/MATH]
[MATH]sd-sb=a^2-c^2[/MATH]
[MATH]s(d-b)=(a+c)(a-c)[/MATH]
And we obtain by subtraction of the first 2 equations:
[MATH](d-b)-(a-c)=r(a-c)[/MATH]
Or:
[MATH]d-b=(r+1)(a-c)[/MATH]
Hence:
[MATH]s(r+1)(a-c)=(a+c)(a-c)[/MATH]
Since \(a\ne c\) (otherwise \(d=b\) and the two quadratics are identical) there results:
[MATH]a+c=s(r+1)[/MATH]
Hence:
[MATH]a+b+c+d=rs(r+1)[/MATH]
We know:
[MATH]c+d=ra[/MATH]
[MATH]a+b=rc[/MATH]
Hence:
[MATH]a+b+c+d=r(a+c)[/MATH]
We also know:
[MATH]c^2-sb=a^2-sd[/MATH]
[MATH]sd-sb=a^2-c^2[/MATH]
[MATH]s(d-b)=(a+c)(a-c)[/MATH]
And we obtain by subtraction of the first 2 equations:
[MATH](d-b)-(a-c)=r(a-c)[/MATH]
Or:
[MATH]d-b=(r+1)(a-c)[/MATH]
Hence:
[MATH]s(r+1)(a-c)=(a+c)(a-c)[/MATH]
Since \(a\ne c\) (otherwise \(d=b\) and the two quadratics are identical) there results:
[MATH]a+c=s(r+1)[/MATH]
Hence:
[MATH]a+b+c+d=rs(r+1)[/MATH]
