We are given that:

$x^2+4x+7=0$

Find the equation with roots $\alpha+2\beta$ and $2\alpha+\beta$

The method which you have to do it in is the simplest method where u=?.

We are NOT allowed to use a brute force method

2. Hello, anthmoo!

We are given that: . $x^2+4x+7\:=\:0$

Find the equation with roots $\alpha+2\beta$ and $2\alpha+\beta$

I assume that $\alpha$ and $\beta$ are the roots of the given equation.

The roots of the quadratic are: . $\begin{array}{cc}\alpha \:=\:\text{-}2 + i\sqrt{3} \\ \beta \:=\:\text{-}2 - i\sqrt{3}\end{array}$

Let $P\:=\:\alpha + 2\beta\:=\:(\text{-}2 + i\sqrt{3}) + 2(\text{-}2 - i\sqrt{3}) \:=\:\text{-}6 - i\sqrt{3}$

Let $Q\:=\:2\alpha + \beta\:=\:2(\text{-}2 + i\sqrt{3}) + (\text{-}2 - i\sqrt{3}) \:=\:\text{-}6 + i\sqrt{3}$

Then: . $P + Q\:=\:(\text{-}6-i\sqrt{3}) + (\text{-}6 + i\sqrt{3}) \:=\:\text{-}12$
. . This is the negative of the x-coefficient.

And: . $PQ\:=\:(\text{-}6-i\sqrt{3})(\text{-}6+i\sqrt{3}) \:=\:39$
. . This is the constant term.

Therefore, the equation is: . $\boxed{x^2 + 12x + 39\:=\:0}$

3. Originally Posted by Soroban
Hello, anthmoo!

I assume that $\alpha$ and $\beta$ are the roots of the given equation.

The roots of the quadratic are: . $\begin{array}{cc}\alpha \:=\:\text{-}2 + i\sqrt{3} \\ \beta \:=\:\text{-}2 - i\sqrt{3}\end{array}$

Let $P\:=\:\alpha + 2\beta\:=\\text{-}2 + i\sqrt{3}) + 2(\text{-}2 - i\sqrt{3}) \:=\:\text{-}6 - i\sqrt{3}" alt="P\:=\:\alpha + 2\beta\:=\\text{-}2 + i\sqrt{3}) + 2(\text{-}2 - i\sqrt{3}) \:=\:\text{-}6 - i\sqrt{3}" />

Let $Q\:=\:2\alpha + \beta\:=\:2(\text{-}2 + i\sqrt{3}) + (\text{-}2 - i\sqrt{3}) \:=\:\text{-}6 + i\sqrt{3}$

Then: . $P + Q\:=\\text{-}6-i\sqrt{3}) + (\text{-}6 + i\sqrt{3}) \:=\:\text{-}12" alt="P + Q\:=\\text{-}6-i\sqrt{3}) + (\text{-}6 + i\sqrt{3}) \:=\:\text{-}12" />
. . This is the negative of the x-coefficient.

And: . $PQ\:=\\text{-}6-i\sqrt{3})(\text{-}6+i\sqrt{3}) \:=\:39" alt="PQ\:=\\text{-}6-i\sqrt{3})(\text{-}6+i\sqrt{3}) \:=\:39" />
. . This is the constant term.

Therefore, the equation is: . $\boxed{x^2 + 12x + 39\:=\:0}$

Thankyou Soroban but it says that I am not allowed to know or work out the roots of the equation. I am only supposed to work it out without knowing the roots.

In another question, it was asked

The equation $x^2+4x+7=0$ has roots $\alpha \ and \ \beta$. By not using a brute force method, find the equations with roots $3\alpha, 3\beta$

I completed this one successfully by using this method simply (the one i need to use):

$Let \ u=3\alpha$
$Therefore \ (1/3)u=\alpha$

This gives me the new equation:

$(\frac{1}{3}u^2)+4(\frac{1}{3}u)+7=0$
$\frac{u^2}{9}+ \frac{4u}{3}+7=0$
$u^2+12u+63=0$

which is the new equation

$x^2+12x+63=0$

Although you got the right answer Soroban, I was strictly not allowed to use such a method! Can you get the same answer using my method?

4. By the way, I still need help with the question I stated in the first post! The second doesn't mean I know how to do the question in the first post

5. Hello again, anthmoo!

I was afraid that used Brute Force . . . I'll try again.

We are given that: $x^2+4x+7=0$

Find the equation with roots $\alpha+2\beta$ and $2\alpha+\beta$

Since $\alpha$ and $\beta$ are roots of the quadratic,
. . we know that: . $\alpha + \beta \:=\:-4$ and $\alpha\beta\:=\:7$

The roots of the new equation are: . $P \:=\:\alpha + 2\beta$ and $Q \:=\:2\alpha + \beta$

The x-coefficient of the new equation is:
. . $-(P + Q) \:=\:-\left[(\alpha + 2\beta) + (2\alpha + \beta)\right] \:=\:-3(\alpha + \beta)$

Since $\alpha + \beta\:=\:-4$, then: . $-(P+Q)\:=\:-3(-4) \:=\:12$
. . Hence, the x-coefficient is: $12$

The constant term of the new equation is:
. . $PQ\:=\:(\alpha + 2\beta)(2\alpha+\beta)$
. . . . . $=\:2\alpha^2 + 5\alpha\beta + 2\beta^2$
. . . . . $= \:2\alpha^2 + 4\alpha\beta + 2\beta^2 + \alpha\beta$
. . . . . $=\:2(\alpha^2 + 2\alpha\beta + \beta^2) + \alpha\beta$
. . . . . $=\:2(\alpha + \beta)^2 + \alpha\beta$

Since $\alpha + \beta\:=\:-4$ and $\alpha\beta\:=\:7$, then: . $PQ\:=\:2(-4)^2 + 7 \:=\:39$
. . Hence, the constant term is $39.$

Therefore, the equation is: . $x^2 + 12x + 39\:=\:0$

6. Ah thankyou VERY much Soroban!

This is due in tomorrow too! Thanks again man!

7. Originally Posted by anthmoo
We are given that:

$x^2+4x+7=0$

Find the equation with roots $\alpha+2\beta$ and $2\alpha+\beta$

The method which you have to do it in is the simplest method where u=?.

We are NOT allowed to use a brute force method

The quadratic with roots $\alpha+2\beta$ and $2\alpha+\beta$ and leading coefficient 1 is:

$x^2-[(\alpha+2\beta)+(2\alpha+\beta)]x + (\alpha+2\beta)(\alpha+2\beta)$,

or after simplifying:

$x^2-(3\alpha+3\beta)x + 2(\alpha + \beta)^2+\alpha \beta$.

But $\alpha$ and $\beta$ are the roots of:

$x^2+4x+7$,

so $\alpha+\beta=-4$, and $\alpha \beta = 7$, so
the quadratic with the required roots is:

$x^2+3\times 4x + 2\times 16+7$,

or:

$x^2+12x + 39$.

RonL

8. Thankyou captainblack!

Hmmm it sounds like CB's way is the way that the book probably wanted it but both ways get the right answer so thanks to both of you.