Help with the math exercise on roots of polynomial equations

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Shas72
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A quadratic equation has roots alpha and beta. Given that 1/ alpha +1/beta=1/2 and alpha^2+beta^2=12, find two possible quadratic equations that satisfy these values.
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Sir Cumference
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(Original post by Shas72)
A quadratic equation has roots alpha and beta. Given that 1/ alpha +1/beta=1/2 and alpha^2+beta^2=12, find two possible quadratic equations that satisfy these values.
What have you tried? Please post your working.
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Shas72
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Yeah I have tried a number of times but can't figure out
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Sir Cumference
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(Original post by Shas72)
Yeah I have tried a number of times but can't figure out
\displaystyle \frac{1}{\alpha} + \frac{1}{\beta}= \frac{1}{2}

First rearrange that so it's in a nicer form. If you've already done this, post your working.
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Shas72
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Shas72
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I dont know if it's the correct way to do. I just saw the answer behind and manipulated
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Shas72
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Shas72
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It's the 8th sum
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Sir Cumference
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(Original post by Shas72)
X
If you rearrange that you have

2(\alpha + \beta) = \alpha \beta


Now using the identity:

(\alpha + \beta)^2 \equiv \alpha^2 + \beta^2 + 2\alpha\beta

You know that \alpha^2+\beta^2 = 12 so that gives two equations:

2(\alpha + \beta) = \alpha \beta

(\alpha + \beta)^2 = 12 + 2\alpha\beta

Any thoughts from here?
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Shas72
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I did that first but couldn't get any solution.
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Sir Cumference
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(Original post by Shas72)
I did that first but couldn't get any solution.
Combining the equations gives

(\alpha+\beta)^2=12+4(\alpha + \beta )

Now this is a quadratic equation - can you see it?
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Shas72
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So how do you get the final two quadratic equations
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Sir Cumference
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(Original post by Shas72)
So how do you get the final two quadratic equations
I'm not sure if you're noticing that this

(\alpha+\beta)^2=12+4(\alpha + \beta )

is a quadratic equation.

If you let X=\alpha + \beta then you have

X^2=12+4X

If you solve this then you have two values for \alpha + \beta which in turn gives you two values for \alpha\beta. This is enough to know what the final quadratic equations are - there's no need to find \alpha and \beta. Make sense?
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Shas72
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I will surely try to do your method. Thanks a tonne for helping.
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Shas72
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(Original post by Notnek)
I'm not sure if you're noticing that this

(\alpha+\beta)^2=12+4(\al\beta )

is a quadratic equation.

If you let X=\alpha + \beta then you have

X^2=12+4X

If you solve this then you have two values for \alpha + \beta which in turn gives you two values for \alpha\beta. This is enough to know what the final quadratic equations are - there's no need to find \alpha and \beta. Make sense?
I couldn't understand how you got 4(alpha+beta)
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Shas72
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(Original post by Shas72)
I couldn't understand how you got 4(alpha+beta)
Oh yeah I got it you substituted it in the second equation.
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Shas72
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How do you get the equations x^2-2x+4 and x^2-6x+12. Do you divide the 4 and 12 from 2
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Shas72
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I got it fully. Thanks thanks thanks a lottttttt. Thanks a tonne for helping me. You are just too good!!!!!! Thanks again
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Shas72
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Can you also pls help me with this one. The quadratic equation 3x^2+2x-4 has roots alpha and beta. Find the values of s1,s2 and s-1.
I finished solving s1 and s1 got the answers s1=-b/a=-2/3 and s2= alpha^2+beta^2= 28/9. I don't know how to find s-1. Pls help
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Shas72
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I got this one. Tried again got this one.
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