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

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

In the cubic equation $x^3 + px^2 + qx + 26 = 0$ the constants p and q are real and $2+3i$ 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 $2-3i$ 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 $\alpha, \beta$ and $\gamma$ in the cubic equation $ax^3 + bx^2 + cx + d$

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

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

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

3. Originally Posted by db5vry
In the cubic equation $x^3 + px^2 + qx + 26 = 0$ the constants p and q are real and $2+3i$ 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 $2-3i$ 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 $(x-\alpha)(x-\beta)(x- \gamma)$ where $\alpha$, $\beta$ and $\gamma$ 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 $(x-2-3i)(x-2+3i)(x-\gamma)$ for some real number $\gamma$. Multiplying, $((x-2)-3i)((x-2)+3i)= (x-2)^2-(3i)^2$ $= (x-2)^2+ 9= x^2- 4x+ 4+ 9= x^2- 4x+ 13$. Now you know that $x^3+ px^2+ qx+ 26= (x^2- 4x+ 13)(x-\gamma)$. I would make a "wild guess" at $\gamma$ being 26/13= 2!.

And once you know that $x^2+ px^2+ qx+ 26= (x^2- 4x+ 13)(x- 2)$ 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 $\alpha, \beta$ and $\gamma$ in the cubic equation $ax^3 + bx^2 + cx + d$

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

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

$\alpha \beta \gamma = \frac{-d}{a}$=-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 $x^3 + px^2 + qx + 26 = 0$ the constants p and q are real and $2+3i$ 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 $2-3i$ 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 $2 + 3i$ is a factor.

So by the remainder and factor theorems, this means that

$f(2 + 3i) = 0$

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

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

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

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

Equating real and imaginary components gives

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

Solve these equations simultaneously.

6. Originally Posted by db5vry
In the cubic equation $x^3 + px^2 + qx + 26 = 0$ the constants p and q are real and $2+3i$ 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 $2-3i$ 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

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

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

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

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

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

$= x^2 - 4x + 13$

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

Then you will have the other factor, and solving this $= 0$ for $x$ 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.