vacinammo288

2022-04-27

Let $\alpha ,\text{}\beta$ be the roots of ${x}^{2}-x+p=0$ and $\gamma ,\text{}\delta$ be the roots of ${x}^{2}-4x+q=0$

where p and q are integers. IfNKS$\alpha ,\text{}\beta ,\text{}\gamma ,\text{}\delta$ are in GP then $p+q$ is?

where p and q are integers. IfNKS

despescarwh9

Beginner2022-04-28Added 15 answers

Step 1

In the first approach: when you solve for r you would have got an equation ${r}^{4}=4$. This gives you ${r}^{2}=2$ and ${r}^{2}=-2$. You have only considered the first case (i.e. when ${r}^{2}=2$). If you consider the second case ${r}^{2}=-2$, then you get $r=\pm i\sqrt{2}$ In which case you will get $p+q=-34$. So both approaches will yield the same answers.

Step 2

Some details:

$\alpha +\beta =1\Rightarrow a\frac{(1+{r}^{2})}{{r}^{3}}=1.$

Likewise

$\gamma +\delta =4\Rightarrow ar(1+{r}^{2})=4.$

Thus

${r}^{4}=4$

This means ${r}^{2}=\pm 2$ Consequently $a=\frac{4}{3r}$ or $a=\frac{-4}{r}$ (the second case occurs when ${r}^{2}=-2$). Now we can use,

$\alpha \cdot \beta =p\Rightarrow p=\frac{{a}^{2}}{{r}^{4}}=\frac{{a}^{2}}{4}.$

Likewise we have

$q={a}^{2}{r}^{4}=4{a}^{2}.$

So,

$p+q=\frac{17}{4}{a}^{2}$

Now plug in the two possible values of $a}^{2}=\frac{16}{9{r}^{2}}=\frac{8}{9$ or ${a}^{2}=\frac{16}{{r}^{2}}=-8$ to get both the possibilities.

Step 3

In fact, from the conditions of the problem that p and q are integers, we can see that ${r}^{2}=2$ is not a valid solution because it gives $a}^{2}=\frac{8}{9$, in which case $p=\frac{{a}^{2}}{4}=\frac{2}{9}\overline{)\in}\mathrm{\mathbb{Z}}$ Thus ${r}^{2}=-2$ is the only valid possibility. In which case

$p+q=-34$

is the only valid answer.

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