Makenna Booker
2022-07-27
Answered

Find integers c and d such that the equation ${x}^{3}+cx+d=0$ has $1+\sqrt{3}$ as one of its roots.

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Anaya Gregory

Answered 2022-07-28
Author has **14** answers

Given an equation ${x}^{3}+b{x}^{2}+cx+d$ that factors to $(x-{r}_{1})(x-{r}_{2})(x-{r}_{3})$,

$b={r}_{1}+{r}_{2}+{r}_{3}$

$c={r}_{1}{r}_{2}+{r}_{2}{r}_{3}+{r}_{3}{r}_{1}$

$d={r}_{1}{r}_{2}{r}_{3}$

For our problem, we'll say that ${r}_{1}=1+\sqrt{3}$

In order for d to be an integer, either ${r}_{2}$ or ${r}_{3}$ must be $(1-\sqrt{3})$ because we have to eliminate theradical in the same way we would if we were rationalizing thedenominator of a fraction.

So far we have ${r}_{1}=1+\sqrt{3},{r}_{2}=1-\sqrt{3}$, and we know that ${r}_{1}+{r}_{2}+{r}_{3}=0$. Using these facts together you should be able to figure out what ${r}_{3}$ is and then plug into the above formul as to find c and d.

$b={r}_{1}+{r}_{2}+{r}_{3}$

$c={r}_{1}{r}_{2}+{r}_{2}{r}_{3}+{r}_{3}{r}_{1}$

$d={r}_{1}{r}_{2}{r}_{3}$

For our problem, we'll say that ${r}_{1}=1+\sqrt{3}$

In order for d to be an integer, either ${r}_{2}$ or ${r}_{3}$ must be $(1-\sqrt{3})$ because we have to eliminate theradical in the same way we would if we were rationalizing thedenominator of a fraction.

So far we have ${r}_{1}=1+\sqrt{3},{r}_{2}=1-\sqrt{3}$, and we know that ${r}_{1}+{r}_{2}+{r}_{3}=0$. Using these facts together you should be able to figure out what ${r}_{3}$ is and then plug into the above formul as to find c and d.

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