Uncovering the Hidden Identity in Solving Quadratic Equation Challenge

[Saitama]

If the quadratic equation $x^2+(2 – \tan \theta)x – (1 + \tan \theta) = 0$ has two integral roots, then sum of all possible values of $\theta$ in the interval $(0, 2\pi)$ is $k\pi$. Find $k$.

 

[anemone]

My solution:

Spoiler

If we rewrite the given equation to make $\tan \theta$ the subject, and then plot a graph of $\tan \theta$ versus $x$, we get:

$\tan \theta=\dfrac{x^2+2x-1}{x+1}$
View attachment 2393
Note:
$x=-1$ is a vertical asymptote

Consider the fact that $x$ are two integrals regardless of what $\tan \theta$ is(i.e. try to imagine we are to move a ruler horizontally on the graph and we have to hit two $x$ of which both are integers), we can deduce that

i. This is impossible for the region where $x<-3$,

ii. This is only possible only when one of the $x$ is in the interval $-3<x<0$.

Therefore

a. when $x=-2$ and this gives $\tan \theta=\dfrac{(-2)^2+2(-2)-1}{(-2)+1}=1$

b. since $x\ne-1$ so we're done.

Hence, the sum of all possible values of $\theta$ in the interval $(0, 2\pi)$ is $\dfrac{\pi}{4}+\dfrac{5\pi}{4}=\dfrac{3\pi}{2}$ and that shows $k=\dfrac{3}{2}$.

 

[Saitama]

My solution:

Spoiler

If we rewrite the given equation to make $\tan \theta$ the subject, and then plot a graph of $\tan \theta$ versus $x$, we get:

$\tan \theta=\dfrac{x^2+2x-1}{x+1}$
View attachment 2393
Note:
$x=-1$ is a vertical asymptote

Consider the fact that $x$ are two integrals regardless of what $\tan \theta$ is(i.e. try to imagine we are to move a ruler horizontally on the graph and we have to hit two $x$ of which both are integers), we can deduce that

i. This is impossible for the region where $x<-3$,

ii. This is only possible only when one of the $x$ is in the interval $-3<x<0$.

Therefore

a. when $x=-2$ and this gives $\tan \theta=\dfrac{(-2)^2+2(-2)-1}{(-2)+1}=1$

b. since $x\ne-1$ so we're done.

Hence, the sum of all possible values of $\theta$ in the interval $(0, 2\pi)$ is $\dfrac{\pi}{4}+\dfrac{5\pi}{4}=\dfrac{3\pi}{2}$ and that shows $k=\dfrac{3}{2}$.

Thank you for your participation anemone but the answer is incorrect. :(

Spoiler

Hint: You have got two correct values for $\theta$, there are still two more. :)

 

[anemone]

Thank you for your participation anemone but the answer is incorrect. :(

It's okay, Pranav, don't be sorry... and I apologize for posting the wrong solution with the incorrect answer for this challenge...

Spoiler

Hint: You have got two correct values for $\theta$, there are still two more. :)

Thanks, I will look more into the problem right NOW, hopefully I can see it...but if I don't, please forgive me!

 

[anemone]

Ah, I see it now:)...

Spoiler

Notice that we need to consider the case when $x=0$ as well and when $x=0$, $\tan \theta=-1$ and that gives the other two values of $\theta=\dfrac{3\pi}{4},\,\dfrac{7\pi}{4}$ in the interval $(0,\,2\pi)$.

Hence, the sum of all possible $\theta$ would be $\dfrac{\pi}{4}+\dfrac{3\pi}{4}+\dfrac{5\pi}{4}+\dfrac{7\pi}{4}=4\pi$, i.e. $k=4$.

 

[Saitama]

Ah, I see it now:)...

Spoiler

Notice that we need to consider the case when $x=0$ as well and when $x=0$, $\tan \theta=-1$ and that gives the other two values of $\theta=\dfrac{3\pi}{4},\,\dfrac{7\pi}{4}$ in the interval $(0,\,2\pi)$.

Hence, the sum of all possible $\theta$ would be $\dfrac{\pi}{4}+\dfrac{3\pi}{4}+\dfrac{5\pi}{4}+\dfrac{7\pi}{4}=4\pi$, i.e. $k=4$.

Excellent!

I wait for others if they want to try the problem before sharing the suggested solution. :)

 

[kaliprasad]

My ans along the line of anemone but algebraically

Spoiler

$\tan \theta = \frac{x^2 + 2x - 1}{x + 1}$
= $(x+1) - \frac{2}{x+1}$
= $y - \frac{2}{y}$ taking x+ 1 = y

for it to be integer we should have y a factor of 2 that is 1 or -1 or 2 or -2

giving $\tan \theta$ = -1 or 1 or 1 or -1

or $\theta = \frac{3\pi}{4}$ or $\frac{\pi}{4}$ or $\frac{5\pi}{4}$ or or $\frac{7\pi}{4}$

adding these we get k = 4

 

[anemone]

Excellent!

I wait for others if they want to try the problem before sharing the suggested solution. :)

Hi Pranav,

If you don't mind me asking, I hope to see the suggested solution or your solution for this problem. As one of regular challenge problems' solvers, I think you would agree with me that it is a "torture" for us not get the chance to see how others approached a challenge problem beautifully, hehehe...:p

 

[Saitama]

Hi Pranav,

If you don't mind me asking, I hope to see the suggested solution or your solution for this problem. As one of regular challenge problems' solvers, I think you would agree with me that it is a "torture" for us not get the chance to see how others approached a challenge problem beautifully, hehehe...:p

It isn't very different from what kaliprasad did.

Spoiler

Let $\alpha$ and $\beta$ be two roots.
Then
$$\alpha+\beta=2-\tan\theta$$
$$\alpha \beta=1+\tan\theta$$
$$\Rightarrow \alpha+\beta+\alpha \beta=3 \Rightarrow (\alpha+1)(\beta+1)=4$$
Now it is easy to find the values of $\alpha$ and $\beta$.

 

[anemone]

It isn't very different from what kaliprasad did.

Spoiler

Let $\alpha$ and $\beta$ be two roots.
Then
$$\alpha+\beta=2-\tan\theta$$
$$\alpha \beta=1+\tan\theta$$
$$\Rightarrow \alpha+\beta+\alpha \beta=3 \Rightarrow (\alpha+1)(\beta+1)=4$$
Now it is easy to find the values of $\alpha$ and $\beta$.

Thanks for your reply!

Ah, I see it now, it is how the identity $x+y+xy+1=(y+1)(x+1)$ that come into play helped! Thanks again, Pranav!