Im not sure how to get to the answers here.

$\alpha+\beta=-4$

$\alpha\beta=3$

$\alpha=$
$\beta=$

would be great to get some help. Thanks.

2. Originally Posted by davidman
Im not sure how to get to the answers here.

$\alpha+\beta=-4$

$\alpha\beta=3$

$\alpha=$
$\beta=$

would be great to get some help. Thanks.
From equation 1 we can see

$\alpha = -\beta - 4$.

Substituting into equation 2 gives

$(-\beta - 4)\beta = 3$

$-\beta^2 - 4\beta = 3$

$0 = \beta^2 + 4\beta + 3$

$0 = (\beta + 3)(\beta + 1)$

$\beta = -3$ or $\beta = -1$.

Now since $\alpha = -\beta - 4$

Case 1:

$\alpha = -(-3) - 4$

$\alpha = -1$.

Case 2:

$\alpha = -(-1) - 4$

$\alpha = -3$.

So the solutions are $\alpha = -1, \beta = -3$ and $\alpha = -3, \beta = -1$.

3. Originally Posted by davidman
Im not sure how to get to the answers here.

$\alpha+\beta=-4$

$\alpha\beta=3$

$\alpha=$
$\beta=$

would be great to get some help. Thanks.
I assume that this is a system of simultaneous equations. If so:

1. Calculate ß from the first equation and plug in this term into the 2nd one:

2. $\alpha(-\alpha-4)=3$

This is a quadratic equation in $\alpha$. Use the quadratic formula to solve this equation.
Spoiler:
$\alpha = -3~\vee~ \alpha = -1$ . Now calculate ß.

EDIT: Too late ...as usual

4. Originally Posted by davidman
Im not sure how to get to the answers here.

$\alpha+\beta=-4$

$\alpha\beta=3$

$\alpha=$
$\beta=$

would be great to get some help. Thanks.
hello,
solve the equation,
$x^2+4x+3$

5. thanks for the really quick input.

I have a similar question where $\alpha^2+\beta^2=32,\alpha\beta=-8$

and it completely throws me off that they're squared all of a sudden.

Do you solve this the same way by taking the square root of LHS and solving the simultaneous equation?

6. Originally Posted by Raoh
hello,
solve the equation,
$x^2+4x+3$
oh I see, $x^2+(\alpha+\beta)x+\alpha\beta$

thanks

7. Originally Posted by davidman
thanks for the really quick input.

I have a similar question where $\alpha^2+\beta^2=32,\alpha\beta=-8$

and it completely throws me off that they're squared all of a sudden.

Do you solve this the same way by taking the square root of LHS and solving the simultaneous equation?
I think the fact that $(\alpha+ \beta)^2= \alpha^2+ 2\alpha\beta+ \beta^2$ will help a lot here!

That will tell you what $\alpha+ \beta$ is (actually, there are two answers to that) and then it becomes just like the previous problem.

8. Originally Posted by davidman
thanks for the really quick input.

I have a similar question where $\alpha^2+\beta^2=32,\alpha\beta=-8$

and it completely throws me off that they're squared all of a sudden.

Do you solve this the same way by taking the square root of LHS and solving the simultaneous equation?
Square the second equation

$\alpha^2\beta^2 = 64$.

Now solve using a similar method to the first question.

9. Originally Posted by davidman
thanks for the really quick input.

I have a similar question where $\alpha^2+\beta^2=32,\alpha\beta=-8$

and it completely throws me off that they're squared all of a sudden.

Do you solve this the same way by taking the square root of LHS and solving the simultaneous equation?
$(\alpha +\beta )^2-2\alpha \beta =32$
$(\alpha +\beta )^2=16$
$\alpha +\beta=4, \text{or}, \alpha +\beta=-4$
now,
solve either $x^2-4x-8$ or $x^2+4x-8$.

10. Originally Posted by Raoh
solve either [...] or $x^2+4x-8$.
ok, I used the quadratic equation on one and it basically turned out like this...

$x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}, a=1, b=4, c=-8$

$x_1=\frac{-4+\sqrt{16+32}}{2}$

$=\frac{-4+4\sqrt{32}}{2}$

$=-2+2\sqrt{32}$

Is this correct?

hmmm... maybe more like

$=\frac{-4+\sqrt{48}}{2}$

$=\frac{-4+\sqrt{4\times{12}}}{2}$

$=-2+\sqrt{12}$

11. Originally Posted by davidman
ok, I used the quadratic equation on one and it basically turned out like this...

$x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}, a=1, b=4, c=-8$

$x_1=\frac{-4+\sqrt{16+32}}{2}$

$=\frac{-4+4\sqrt{32}}{2}$

$=-2+2\sqrt{32}$

Is this correct?
$x_1=-2(1+\sqrt{3})$
and $x_2=2(\sqrt{3}-1)$.
P.S : $\sqrt{16+32}=4\sqrt{3}$.
$\sqrt{16+32}=\sqrt{48}=\sqrt{4\times4\times3}$