Let $α,β$ be the roots of $x^2-x+p=0$ and $γ,δ$ be the roots of $x^2-4x+q = 0$ where $p$ and $q$ are integers. If $α,β,γ,δ$ are in GP then $p + q$ is?

Question

Let $α,β$ be the roots of $x^2-x+p=0$ and $γ,δ$ be the roots of $x^2-4x+q = 0$ where $p$ and $q$ are integers. If $α,β,γ,δ$ are in GP then $p + q$ is ?

My solution approach :-
I have assumed $α,β,γ,δ$ to be as follows ;
$α = \frac{a}{r^3}$
$β = \frac{a}{r}$
$γ = ar$
$δ = ar^3$
Here the common ratio is $r^2$. Now $α+β = 1$ and $γ+δ = 4$ and solving these two equations in terms of $a$ and $r$ gives the value of $r= \sqrt{2} \text{ and } \sqrt{-2}$ which leads to the values of $a$ to be $\frac{2\sqrt{2}}{3}$ and $\frac{-2\sqrt{2}}{3}$ respectively.
In both the above cases, $α,β,γ,δ$ turns out to be the same i.e. $\frac{1}{3},\frac{2}{3},\frac{4}{3},\frac{8}{3}$ which gives the value of $p+q = \frac{34}{9}$.

Now if I assume $α,β,γ,δ$ to be as follows ;
$α = a$
$β = ar$
$γ = ar^2$
$δ = ar^3$
Here the common ratio is just $r$. But in this case $r$ values turn out to be $\pm2$ and for $r=2$ we get $α,β,γ,δ$ to be $\frac{1}{3},\frac{2}{3},\frac{4}{3},\frac{8}{3}$ whereas for $r=-2$ we get $α,β,γ,δ$ to be $-1,2,-4,8$ and hence $p+q = \frac{34}{9} \text{ or }-34 $.

Now I am getting confused why there is this difference in both the approaches. What am I doing wrong? Please help me on this !!!

Thanks in advance !!!

Answer 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.

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Some details: $$\alpha+\beta=1 \implies a\frac{(1+r^2)}{r^3}=1.$$ Likewise $$\gamma+\delta=4 \implies ar(1+r^2)=4.$$ 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, $$\alpha \cdot \beta=p \implies p=\frac{a^2}{r^4}=\frac{a^2}{4}.$$ Likewise we have $$q=a^2r^4=4a^2.$$ So, $$p+q=\frac{17}{4}a^2$$ Now plug in the two possible values of $a^2=\frac{16}{9r^2}=\frac{8}{9}$ OR $a^2=\frac{16}{r^2}=-8$ to get both the possibilities.

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Additional remark:

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 $p=\frac{a^2}{4}=\frac{2}{9} \not\in \Bbb{Z}$. Thus $\color{red}{r^2=-2}$ is the ONLY valid possibility. In which case $$\color{magenta}{p+q=-34}$$ is the ONLY valid answer.

Answer 2

The question can also be approached from the direction of working with the zeroes of the polynomials themselves. For $ \ x^2 - x + p \ = \ 0 \ $ and $ \ x^2 - 4x + q \ = \ 0 \ \ , $ we have $$ \alpha \ , \ \beta \ \ = \ \ \frac{1 \pm \sqrt{1 - 4p}}{2} \ = \ \frac12 \ \pm \ \frac12 \sqrt{1-4p} \ = \ \frac12 \ \pm \ \frac12\sqrt{D_1} \ \ , $$ $$ \gamma \ , \ \delta \ \ = \ \ \frac{4 \pm \sqrt{16 - 4q}}{2} \ = \ 2 \ \pm \ \sqrt{4-q} \ = \ 2 \ \pm \ \sqrt{D_2} \ \ . $$

We wish for the zeroes to lie in a geometric progression $ \ \alpha \ , \ \beta \ , \ \gamma \ , \ \delta \ \ , $ which for real zeroes requires:

• either a ratio $ \ r > 0 \ $ with all of the zeroes positive and in the indicated sequence, or

• a ratio $ \ r < 0 \ $ with the zeroes alternating in sign and falling on the real-number line in the sequence $ \ \gamma \ , \ \alpha \ , \ (0) \ , \ \beta \ , \ \delta \ \ . $

[Note: we are using $ \ r \ $ as the ratio between zeroes here, rather than $ \ r^2 \ . $ ]

We can see already what is necessary for the first case: if $ \ \alpha \ > 0 \ $ (and likewise for the other zeroes) and the zeroes are all to be distinct, we must have $ \ 0 < 1-4p < 1 \ \Rightarrow \ 0 < p < \frac14 \ \ . $ Thus, if $ \ p \ $ and $ \ q \ $ are to be integers, the first case above has no solution. (We will relax this condition later to see what occurs.) On the other hand, if we permit $ \ \alpha < 0 \ $ with the other zeroes having alternating signs, we simply have $ \ 1 < 1-4p \ \Rightarrow \ p < 0 \ \ . $ (We can also find constraints for $ \ q \ $ here, but we'll shortly see that a quite rigid requirement will be set.)

If we first consider the required ratio for each pair of zeroes, we find $$ \frac12 · ( \ 1 + \sqrt{D_1} \ ) \ = \ r · \frac12 · ( \ 1 - \sqrt{D_1} \ ) \ \ \Rightarrow \ \ D_1 \ = \ \left( \frac{r \ - \ 1}{r \ + \ 1} \right)^2 \ \ , $$ $$ 2 + \sqrt{D_2} \ = \ r · ( \ 2 - \sqrt{D_2} \ ) \ \ \Rightarrow \ \ D_2 \ = \ \left( \frac{2r \ - \ 2}{r \ + \ 1} \right)^2 \ = \ 4 · \left( \frac{r \ - \ 1}{r \ + \ 1} \right)^2 \ = \ 4D_1 \ \ . $$

From this, we obtain $ \ 4 - q \ = \ 4 · (1-4p) \ \Rightarrow \ q \ = \ 16p \ \ , $ and consequently that $ \ p + q \ = \ 17p \ \ . $

We now need to "couple" these pairs of zeroes through the condition $ \ \gamma \ = \ r · \beta \ $ or $$ 2 - \sqrt{D_2} \ = \ r · \frac12 · ( \ 1 + \sqrt{D_1} \ ) \ \ \Rightarrow \ \ 2·2 \ - \ 2·2\sqrt{D_1} \ = \ r · ( \ 1 + \sqrt{D_1} \ ) $$ $$ \Rightarrow \ \ D_1 \ = \ \left( \frac{4 \ - \ r}{r \ + \ 4} \right)^2 \ \ . $$

Equating the two ratio expressions for $ \ D_1 \ $ yields either

$$ \left( \frac{r \ - \ 1}{r \ + \ 1} \right) \ = \ \left( \frac{4 \ - \ r}{r \ + \ 4} \right) \ \ \ \text{or} \ \ \ \left( \frac{r \ - \ 1}{r \ + \ 1} \right) \ = \ \left( \frac{r \ - \ 4}{r \ + \ 4} \right) \ \ ; $$

the first of these leads to $ \ r^2 + 3r - 4 \ = \ -r^2 + 3r + 4 \ \Rightarrow \ 2r^2 \ = \ 8 \ \Rightarrow \ r^2 \ = \ 4 \ \ , $ while the second equation produces $ \ r^2 + 3r - 4 \ = \ r^2 - 3r - 4 \ \Rightarrow \ 6r \ = \ 0 \ , $ which is a "trivial" result.

The first of the cases (for which we already know $ \ p \ $ and $ \ q \ $ cannot be integral) gives us

$ \mathbf{ r = 2 \ \ : } \quad D_1 \ = \ \left( \frac{2 \ - \ 1}{2 \ + \ 1} \right)^2 \ = \ \frac19 \ = \ 1 - 4p \ \ \Rightarrow \ \ p \ = \ \frac29 \ \ \Rightarrow \ \ p+q \ = \ \frac{34}{9} \ \ $

[we do have a rational-number solution] with the zeroes being $ \ \alpha \ , \ \beta \ = \ \frac12 \ \pm \ \frac12\sqrt{\frac19} \ = \ \frac13 \ , \ \frac23 \ $ and $ \ \gamma \ , \ \delta \ = \ 2 \ \pm \ \sqrt{4·\frac19} \ = \ \frac43 \ , \ \frac83 \ \ . $

The alternating-sign case is

$ \mathbf{ r = -2 \ \ : } \quad D_1 \ = \ \left( \frac{[-2] \ - \ 1}{[-2] \ + \ 1} \right)^2 \ = \ 9 \ = \ 1 - 4p \ \ \Rightarrow \ \ p \ = \ -2 \ \ \Rightarrow \ \ p+q \ = \ -34 \ \ $

with the zeroes being $ \ \alpha \ , \ \beta \ = \ \frac12 \ \pm \ \frac12·\sqrt{9} \ = \ -1 \ , \ 2 \ $ and $ \ \gamma \ , \ \delta \ = \ 2 \ \pm \ \sqrt{4·9} \ = \ -4 \ , \ 8 \ \ , $ arranged on the $ \ x-$axis in the order $ \ \gamma \ , \ \alpha \ , \ \beta \ , \ \delta \ \ . $

Graphs for the two cases are presented below.

$ \ \ $

We won't need to be concerned about the situation for complex zeroes. We would then have $$ \alpha \ , \ \beta = \overline{\alpha} \ \ = \ \ \frac12 \ \pm \ i · \frac12 \sqrt{4p-1} \ \ , \ \ \gamma \ , \ \delta = \overline{\gamma} \ \ = \ \ 2 \ \pm \ i · \sqrt{q-4} \ \ , \ \ p > \frac14 \ , \ q > 4 \ \ . $$ Since these are conjugate-pair zeroes which may be expressed as $ \ \zeta · e^{\pm i \phi} \ $ and $ \ \xi · e^{\pm i \psi} \ \ , $ respectively, there is no single complex ratio $ \ r \ = \ \rho · e^{i \theta} \ \ , $ with $ \ \rho \ , \ \zeta \ $ and $ \ \xi \ $ being real numbers, that will put all four zeroes into a geometric progression.