CAT 2022 Slot 2 · Quant · Question 9

A cubic has three roots, so call the third one s alongside the given r and -r.

For 5{x}^{3}+c{x}^{2}-10x+9=0, Vieta's relations compare each coefficient with 5.

The sum of the roots is -\frac{c}{5}, and here that sum is r+\left( -r \right)+s=s.

So s=-\frac{c}{5}.

The sum of the products of roots taken two at a time is \frac{-10}{5}=-2.

Those products are r\left( -r \right)+rs+\left( -r \right)s=-{r}^{2}, since the last two cancel.

So -{r}^{2}=-2, giving {r}^{2}=2.

The product of all three roots is -\frac{9}{5}.

That product is r\left( -r \right)s=-{r}^{2}s=-2s.

So -2s=-\frac{9}{5}, giving s=\frac{9}{10}.

Finally s=-\frac{c}{5} gives c=-5s=-5\times \frac{9}{10}=-\frac{9}{2} — the first option.

Shortcut. Roots r and -r make the cubic's odd and even parts split: substituting both and subtracting kills the even terms, leaving 5{r}^{3}-10r=0, so {r}^{2}=2. Adding instead gives c{r}^{2}+9=0, so c=-\frac{9}{2} in two lines.