If $c,d$ are the real roots of the equation, $(x-a)(x-b)=f$

Question

EDIT:

Sorry, corrected the Typo

If $c,d$ are the real roots of the equation, $$(x-a)(x-b)=f$$ Then roots of the equation, $$(x-c)(x-d)+f=0$$ are?

1. $a,b$
2. $\frac{a}{f},\frac{b}{f}$
3. $\frac{f}{a},\frac{f}{b}$
4. $\frac{c}{f},\frac{d}{f}$

I don't have any hints on how to proceed with this. The only thing I was able to establish was, $$a+b=c+d$$

$$ab=cd+f$$

Any hints would be helpful. Thank you.

Answer 1

The answer can be achieved by using the same technique again.

That is, consider the roots of the equation $$(x-a)(x-b)-f = x^2 - (a+b)x +(ab-f) = 0$$

As the OP has already realised, the sum of the roots of this quadratic are $(a+b)$, and the product is $(ab-f)$.

Given that the roots are $c,d$, we can thus conclude that $$ c+d = a+b \quad cd = ab-f $$

We now simply apply this same logic to the second equation. $$ (x-c)(x-d) + f = 0 \implies x^2-(c+d)x +(cd-f) = 0$$

So the sum of the roots of this equation are $c+d$ and the product of the roots are ($cd-f)$.

But we have already determined that $$ c+d = a+b \quad cd = ab-f $$

Thus, the sum roots of the second equation are $a+b$, and the product is $ab$.

Thus it is clear that the answer is option 1: the roots are $a,b$

Answer 2

Trusting that $(x-c)(x-d)+f=0$ was intended, just remark that $$(x-a)(x-b)-f=(x-c)(x-d)\implies (x-c)(x-d)+f=(x-a)(x-b)$$

Easy to see that this generalizes to higher degree.