Im not sure how to get to the answers here.

$\displaystyle \alpha+\beta=-4$

$\displaystyle \alpha\beta=3$

$\displaystyle \alpha=$

$\displaystyle \beta=$

would be great to get some help. Thanks.

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- Jan 21st 2010, 04:20 AM #1

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- Jan 21st 2010, 04:27 AM #2
From equation 1 we can see

$\displaystyle \alpha = -\beta - 4$.

Substituting into equation 2 gives

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

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

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

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

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

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

Case 1:

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

$\displaystyle \alpha = -1$.

Case 2:

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

$\displaystyle \alpha = -3$.

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

- Jan 21st 2010, 04:27 AM #3
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. $\displaystyle \alpha(-\alpha-4)=3$

This is a quadratic equation in $\displaystyle \alpha$. Use the quadratic formula to solve this equation.

__Spoiler__:

EDIT: Too late ...as usual

- Jan 21st 2010, 04:31 AM #4

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- Jan 21st 2010, 04:34 AM #5

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thanks for the really quick input.

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

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?

- Jan 21st 2010, 04:35 AM #6

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- Jan 21st 2010, 04:37 AM #7

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- Jan 21st 2010, 04:37 AM #8

- Jan 21st 2010, 04:44 AM #9

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- Jan 21st 2010, 05:02 AM #10

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ok, I used the quadratic equation on one and it basically turned out like this...

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

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

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

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

Is this correct?

hmmm... maybe more like

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

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

$\displaystyle =-2+\sqrt{12}$

- Jan 21st 2010, 05:26 AM #11

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- Jan 21st 2010, 05:32 AM #12

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