# Thread: How is it possible to find the other root?

1. ## How is it possible to find the other root?

In the cubic equation $\displaystyle x^3 + px^2 + qx + 26 = 0$ the constants p and q are real and $\displaystyle 2+3i$ is a root. I want to find the other two roots but I'm a little lost at what to do to get one of them.

I know that $\displaystyle 2-3i$ is a root as the conjugate is always a root, and I also know that a cubic equation has either one or three real roots - so it follows that the other root I need to find is real.

I tried finding the sum and product of the roots but I can't because I don't know what p and q are yet - I tried using an algebraic division method but the same can be said for this, and I don't know any other method to get the other root.

The second part of the question asks to find the values of p and q, but I take it that there's a way to get the other root without getting these first? Could anyone suggest a method?

Thanks if you can help me out

2. Have you tried the sum and products of roots?

Given roots $\displaystyle \alpha, \beta$ and $\displaystyle \gamma$ in the cubic equation $\displaystyle ax^3 + bx^2 + cx + d$

$\displaystyle \alpha + \beta + \gamma = \frac{-b}{a}$

$\displaystyle \alpha \beta + \alpha \gamma + \beta \gamma = \frac{c}{a}$

$\displaystyle \alpha \beta \gamma = \frac{-d}{a}$

3. Originally Posted by db5vry
In the cubic equation $\displaystyle x^3 + px^2 + qx + 26 = 0$ the constants p and q are real and $\displaystyle 2+3i$ is a root. I want to find the other two roots but I'm a little lost at what to do to get one of them.

I know that $\displaystyle 2-3i$ is a root as the conjugate is always a root, and I also know that a cubic equation has either one or three real roots - so it follows that the other root I need to find is real.
Also, I presume you know that a cubic, with leading coefficient 1, is equal to $\displaystyle (x-\alpha)(x-\beta)(x- \gamma)$ where $\displaystyle \alpha$, $\displaystyle \beta$ and $\displaystyle \gamma$ are the roots. You know that one root is 2+ 3i and another is 2- 3i. That tells you that this can be written as $\displaystyle (x-2-3i)(x-2+3i)(x-\gamma)$ for some real number $\displaystyle \gamma$. Multiplying, $\displaystyle ((x-2)-3i)((x-2)+3i)= (x-2)^2-(3i)^2$$\displaystyle = (x-2)^2+ 9= x^2- 4x+ 4+ 9= x^2- 4x+ 13$. Now you know that $\displaystyle x^3+ px^2+ qx+ 26= (x^2- 4x+ 13)(x-\gamma)$. I would make a "wild guess" at $\displaystyle \gamma$ being 26/13= 2!.

And once you know that $\displaystyle x^2+ px^2+ qx+ 26= (x^2- 4x+ 13)(x- 2)$ it is easy to find p and q.

I tried finding the sum and product of the roots but I can't because I don't know what p and q are yet - I tried using an algebraic division method but the same can be said for this, and I don't know any other method to get the other root.

The second part of the question asks to find the values of p and q, but I take it that there's a way to get the other root without getting these first? Could anyone suggest a method?

Thanks if you can help me out

4. Originally Posted by Gusbob
Have you tried the sum and products of roots?

Given roots $\displaystyle \alpha, \beta$ and $\displaystyle \gamma$ in the cubic equation $\displaystyle ax^3 + bx^2 + cx + d$

$\displaystyle \alpha + \beta + \gamma = \frac{-b}{a}$ = -p

$\displaystyle \alpha \beta + \alpha \gamma + \beta \gamma = \frac{c}{a}$ = q

$\displaystyle \alpha \beta \gamma = \frac{-d}{a}$=-26
I really don't see that going anywhere. Could you find a way to get the real root from the two that are already known?

5. Originally Posted by db5vry
In the cubic equation $\displaystyle x^3 + px^2 + qx + 26 = 0$ the constants p and q are real and $\displaystyle 2+3i$ is a root. I want to find the other two roots but I'm a little lost at what to do to get one of them.

I know that $\displaystyle 2-3i$ is a root as the conjugate is always a root, and I also know that a cubic equation has either one or three real roots - so it follows that the other root I need to find is real.

I tried finding the sum and product of the roots but I can't because I don't know what p and q are yet - I tried using an algebraic division method but the same can be said for this, and I don't know any other method to get the other root.

The second part of the question asks to find the values of p and q, but I take it that there's a way to get the other root without getting these first? Could anyone suggest a method?

Thanks if you can help me out
You know that $\displaystyle 2 + 3i$ is a factor.

So by the remainder and factor theorems, this means that

$\displaystyle f(2 + 3i) = 0$

So $\displaystyle (2 + 3i)^3 + p(2 + 3i)^2 + q(2 + 3i) + 26 = 0$

$\displaystyle 8 + 36i - 54 - 27i + p(4 + 12i - 9) + 2q + 3qi + 26 = 0$

$\displaystyle -20 + 9i - 5p + 12pi + 2q + 3qi = 0$

$\displaystyle 2q - 5p + i(12p + 3q) = 20 + 9i$

Equating real and imaginary components gives

$\displaystyle 2q - 5p = 20$ and $\displaystyle 12p + 3q = 9$.

Solve these equations simultaneously.

6. Originally Posted by db5vry
In the cubic equation $\displaystyle x^3 + px^2 + qx + 26 = 0$ the constants p and q are real and $\displaystyle 2+3i$ is a root. I want to find the other two roots but I'm a little lost at what to do to get one of them.

I know that $\displaystyle 2-3i$ is a root as the conjugate is always a root, and I also know that a cubic equation has either one or three real roots - so it follows that the other root I need to find is real.

I tried finding the sum and product of the roots but I can't because I don't know what p and q are yet - I tried using an algebraic division method but the same can be said for this, and I don't know any other method to get the other root.

The second part of the question asks to find the values of p and q, but I take it that there's a way to get the other root without getting these first? Could anyone suggest a method?

Thanks if you can help me out
To find the third root, you know that

$\displaystyle x - (2 + 3i)$ and $\displaystyle x - (2 - 3i)$ are factors of the cubic.

Therefore $\displaystyle [x - (2 + 3i)][x - (2 - 3i)]$ is also a factor.

$\displaystyle =(x - 2 - 3i)(x - 2 + 3i)$

$\displaystyle = x(x - 2 + 3i) - 2(x - 2 + 3i) - 3i(x - 2 + 3i)$

$\displaystyle = x^2 - 2x + 3xi - 2x + 4 - 6i - 3xi + 6i + 9$

$\displaystyle = x^2 - 4x + 13$

So now divide your cubic (and from my previous post you will have found $\displaystyle p$ and $\displaystyle q$) by $\displaystyle x^2 - 4x + 13$.

Then you will have the other factor, and solving this $\displaystyle = 0$ for $\displaystyle x$ will give you the third root.

7. Originally Posted by db5vry
I really don't see that going anywhere. Could you find a way to get the real root from the two that are already known?
Post #3 told you exactly what to do and then did it for you.