Operations on zeroes of quadratic equations

Question

If $a$ and $b$ are the zeroes of $f(x) =x^2+3x+1$ and and $c$ and $d$ are the zeroes of $g(x) = x^2+ 4x+1$, then evaluate

$$E = \dfrac{(a-c)(b-c)(a+d)(b+d)}{2}$$

I could find the roots of the above two equations. However, the value of $c$ has be subtracted and the value of $d$ has to be added. How do I decide which one is $c$ and which one should be taken as $d$?

Answer 1

We have: $$(a-c)(b-c) =(c^2-(a+b)c+ab)$$ Now, we know that if $\alpha$ and $\beta$ are the roots of a quadratic equation, $z^2-az+b=0$, then: $$\alpha + \beta = a\, ; \alpha\beta = b$$ Using this fact here, we have: $a+b = -3$ and $ab =1$, as $a$ and $b$ are the roots of $x^2+3x+1=0$.

Thus, $(a-c)(b-c) = (c^2+3c+1) = (1+3c -4c-1)=-c$ , because $c$ is a root of the quadratic $x^2+4x+1=0$.

Similarly, $(a+d)(b+d) = -7d$, giving us: $$\frac{(a-c)(b-c)(a+d)(b+d)}{2}= \frac{7}{2}$$

Answer 2

Use Vieta's relations: \begin{align} (a-c)(b-c)(a+d)(b+d)&=\bigl(ab-(a+b)c+c^2\bigr)\bigl(ab+(a+b)d+d^2\bigr)\\ &=(1+3c+c^2)(1-3d+d^2)\\ &=(1+3c-4c-1)(1-3d-4d-1)\\ &=7cd=7. \end{align}

Answer 3

You don't have to worry.

Observe that

$$f(c)=g(c)-c=0-c$$

and

$$f(-d)=g(d)-7d=0-7d.$$

Then

$$2E=f(c)f(-d)=7cd$$ and you can freely swap the roots.

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As a bonus, Vieta tells us that $cd=1$.