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Quadratic equation

Posted June 8th, 2009 by alpha
in

If $ \alpha $ and $ \beta $ are the roots of $ x^2+px-q=0 $ and$ \gamma $ and $ \delta $ are the roots of $ x^2+px+r=0 $ show that $ (\alpha-\gamma)(\alpha-\delta)=(\beta-\gamma)(\beta-\delta)=q+r $

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answer

On June 8th, 2009 Structure says:

I have solved the problem.Both expressions are equal to q+r

answer

On June 8th, 2009 Structure says:

Prove that :$ (\alpha-\gamma)(\alpha-\delta)=(\beta-\gamma)(\beta-\delta)=q+r $
$ \alpha+\beta=-p , \alpha.\beta=-q $
$ \gamma+\delta=-p , \gamma.\delta=r $

$ (\alpha-\gamma)(\alpha-\delta) $=$ \alpha^2-\alpha(\delta+\gamma)+\gamma\delta= \alpha^2-\alpha(\alpha+\beta)+\gamma\delta= $
=$ \alpha^2-\alpha^2-\alpha\beta+r=q+r $ .....................(1)

similarly
$ (\beta-\gamma) (\beta-\delta) $=$ \beta^2-\beta(\gamma+\delta)\delta\gamma =\beta^2-\beta(\alpha+\beta)+\gamma\delta= $
=$ \beta^2-\beta^2-\alpha\beta+r $=$ q+r $ ....................(2)

Please check my solution and

On June 8th, 2009 alpha says:

Please check my solution and guide me how to continue.I don't follow equations (1)&(2) in ur solution.
So,please explain those steps.

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working

On June 8th, 2009 alpha says:

Prove that :$ (\alpha-\gamma)(\alpha-\delta)=(\beta-\gamma)(\beta-\delta)=q+r $
$ \alpha+\beta=-p , \alpha.\beta=-q $
$ \gamma+\delta=-p , \gamma.\delta=r $

$ (\alpha-\gamma)(\alpha-\delta) $=$ \alpha^2-\alpha(\delta+\gamma)+\gamma\delta $
=$ \alpha^2-\alpha(-p)+r=\alpha^2+p\alpha+r $ .....................(1)

similarly
$ (\beta-\gamma) (\beta-\delta) $=$ \beta^2-\beta(\gamma+\delta)\delta\gamma $
=$ \beta^2-\beta(-p)+r $=$ \beta^2+\betap+r $ ....................(2)

By observing equation(1) , equation(2)&$ x^2+px+r $
$ \alpha $ and $ \beta $ are the roots of the equation$ x^2+px+r $

I have reached upto this.
How to continue?

````Maths````
000
(((())))
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