Thursday, July 19, 2018

Find the sum \(a+b+c+d\).

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.

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]