# Thread: Range of a specific value in quadratic equation to have 3 real roots

1. ## Range of a specific value in quadratic equation to have 3 real roots

Can anyone help me with the this?

Shenaya Perera

2. ## Re: Range of a specific value in quadratic equation to have 3 real roots

Hey pererashenaya7.

It means you have to find a value of k where you can get three solutions for x to satisfy f(x) = 0 which you will have to define.

3. ## Re: Range of a specific value in quadratic equation to have 3 real roots

Originally Posted by chiro
Hey pererashenaya7.

It means you have to find a value of k where you can get three solutions for x to satisfy f(x) = 0 which you will have to define.
Yes...I know that...:-)

For example if the discriminant of a quadratic equation is greater than or equal to 0 it has 2 real roots and if it is equal to zero it has one real root.
But in here we have to find the range of k it has three real roots.But how? And that's exactly where I've been stuck... :-)

Shenaya Perera

4. ## Re: Range of a specific value in quadratic equation to have 3 real roots

Let y= tan(x). Then the equation is $y^2- y- k= 0$. That has two distinct real roots if and only if $(-1)^2- 4(1)(-k)= 4k+ 1\ge 0$. Now, what about tan(x)? For a given value of y, how many roots does $tan(x)= k$ have?

5. ## Re: Range of a specific value in quadratic equation to have 3 real roots

Originally Posted by HallsofIvy
Let y= tan(x). Then the equation is $y^2- y- k= 0$. That has two distinct real roots if and only if $(-1)^2- 4(1)(-k)= 4k+ 1\ge 0$. Now, what about tan(x)? For a given value of y, how many roots does $tan(x)= k$ have?
Two real roots...

But the question asks for a range of k which the equation has three real roots??.That's where I've been stuck..

Shenaya Perera

6. ## Re: Range of a specific value in quadratic equation to have 3 real roots

Maybe I'm missing something in the question (see attached graph of $y=\tan^2{x}-\tan{x}$) ...

For $k < -\dfrac{1}{4}$, $\tan^2{x}-\tan{x} = k$ has no real solutions.

For $k \ge -\dfrac{1}{4}$, $\tan^2{x}-\tan{x}=k$ has an infinite number of real solutions.

If we're considering only a single period of the function ...

$k < -\dfrac{1}{4}$ yields no solutions, $k = -\dfrac{1}{4}$ yields a single solution, and $k > - \dfrac{1}{4}$ yields only two solutions.

Sorry, but I'm not seeing how the equation yields three real solutions unless there is a restricted interval for $x$ that was unstated in the original problem ... someone please enlighten me.

7. ## Re: Range of a specific value in quadratic equation to have 3 real roots

Originally Posted by skeeter
Maybe I'm missing something in the question (see attached graph of $y=\tan^2{x}-\tan{x}$) ...

For $k < -\dfrac{1}{4}$, $\tan^2{x}-\tan{x} = k$ has no real solutions.

For $k \ge -\dfrac{1}{4}$, $\tan^2{x}-\tan{x}=k$ has an infinite number of real solutions.

If we're considering only a single period of the function ...

$k < -\dfrac{1}{4}$ yields no solutions, $k = -\dfrac{1}{4}$ yields a single solution, and $k > - \dfrac{1}{4}$ yields only two solutions.

Sorry, but I'm not seeing how the equation yields three real solutions unless there is a restricted interval for $x$ that was unstated in the original problem ... someone please enlighten me.
No you aren't missing anything...I've posted the exact question..
I'm also stuck on the same case,I also don't see a way to find a range of k which this equation has 3 real roots..

Shenaya Perera

8. ## Re: Range of a specific value in quadratic equation to have 3 real roots

Originally Posted by pererashenaya7
No you aren't missing anything...I've posted the exact question..
I'm also stuck on the same case,I also don't see a way to find a range of k which this equation has 3 real roots..

Shenaya Perera
... then I advise you research the origin of this question or ask whoever assigned it for clarification.