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Help with the math exercise on roots of polynomial equations

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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Reply 1
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.
Reply 2
Yeah I have tried a number of times but can't figure out
Reply 3
Original post by Shas72
Yeah I have tried a number of times but can't figure out

1α+1β=12\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.
Reply 4
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Reply 5
I dont know if it's the correct way to do. I just saw the answer behind and manipulated
Reply 6
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Reply 7
It's the 8th sum
Reply 8
Original post by Shas72
X

If you rearrange that you have

2(α+β)=αβ2(\alpha + \beta) = \alpha \beta


Now using the identity:

(α+β)2α2+β2+2αβ(\alpha + \beta)^2 \equiv \alpha^2 + \beta^2 + 2\alpha\beta

You know that α2+β2=12\alpha^2+\beta^2 = 12 so that gives two equations:

2(α+β)=αβ2(\alpha + \beta) = \alpha \beta

(α+β)2=12+2αβ(\alpha + \beta)^2 = 12 + 2\alpha\beta

Any thoughts from here?
Reply 9
I did that first but couldn't get any solution.
Reply 10
Original post by Shas72
I did that first but couldn't get any solution.

Combining the equations gives

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

Now this is a quadratic equation - can you see it?
Reply 11
So how do you get the final two quadratic equations
Reply 12
Original post by Shas72
So how do you get the final two quadratic equations

I'm not sure if you're noticing that this

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

is a quadratic equation.

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

X2=12+4XX^2=12+4X

If you solve this then you have two values for α+β\alpha + \beta which in turn gives you two values for αβ\alpha\beta. This is enough to know what the final quadratic equations are - there's no need to find α\alpha and β\beta. Make sense?
Reply 13
I will surely try to do your method. Thanks a tonne for helping.
Reply 14
Original post by Notnek
I'm not sure if you're noticing that this

Unparseable latex formula:

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



is a quadratic equation.

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

X2=12+4XX^2=12+4X

If you solve this then you have two values for α+β\alpha + \beta which in turn gives you two values for αβ\alpha\beta. This is enough to know what the final quadratic equations are - there's no need to find α\alpha and β\beta. Make sense?

I couldn't understand how you got 4(alpha+beta)
Reply 15
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.
Reply 16
How do you get the equations x^2-2x+4 and x^2-6x+12. Do you divide the 4 and 12 from 2
Reply 17
I got it fully. Thanks thanks thanks a lottttttt. Thanks a tonne for helping me. You are just too good!!!!!! Thanks again
Reply 18
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
Reply 19
I got this one. Tried again got this one.

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