mathslover:
Tutorial Drill #2 Quadratic Equations
mathslover:
So , Today i am going to post a tutorial on A theorem on quadratic equations : Have you ever thought of that is their any proof of : Can a quadratic equation only have 2 roots , cant it more ? There is a simple reason that is the degree of polynomial is 2 . But I am going to post the exact proof for this : (* 2 distinct real roots ) Quadratic equation general formula
> \(ax^2 + bx + c = 0\) So let us have : three roots : α , β , γ : alpha , beta and gamma are distinct roots and none of them equals zero . If alpha , beta and gamma are the roots then it is necessary that: \(a\alpha^2 + b \alpha +c=0 \)
(1) \(a \beta^2 + b \beta + c = 0\)
(2) \(a \gamma^2 + b \gamma + c =0\)
(3) So first of all we will subtract eq. 2 from eq. 1 that is : \(\large{a(\alpha^2)+b(\alpha)+c-a(\beta^2)-b(\beta)-c=0}\) \(\large{a(\alpha^2)-a(\beta^2)+b(\alpha)-b(\beta)+c-c=0}\) \(\large{a(\alpha^2-\beta^2)+b(\alpha-\beta)=0}\) \(\large{a(\alpha+\beta)(\alpha-\beta)+b(\alpha-\beta)=0}\) Applying inverse of distributive property we get : \(\large{(\alpha-\beta)[a(\alpha+\beta)+b]=0}\) \(\large{a(\alpha+\beta)+b=0}\)
(4) SINCE : we have taken alpha and beta as two distinct roots , hence their diff. can not be 0 . Now similarly subtracting eq. (3) from eq. (2) \(\large{a(\gamma^2)+b(\gamma)+c-a(\beta^2)-b(\beta)-c=0}\) \(\large{a(\gamma^2)-a(\beta^2)+b(\gamma)-b(\beta)+c-c=0}\) \(\large{a(\gamma^2-\beta^2)+b(\gamma-\beta)=0}\) \(\large{a(\gamma+\beta)(\gamma-\beta)+b(\gamma-\beta)=0}\) \(\large{(\gamma-\beta)[a(\gamma+\beta)+b] = 0}\) \(\large{a(\gamma+\beta)+b=0}\)
(5) SINCE : We have taken gamma and beta as two distinct roots , hence their diff. can not be 0 Now Subtracting eq.(5) from eq.(4) : \(\large{a(\alpha+\beta)+b-a(\gamma+\beta)-b=0}\) \(\large{a(\alpha+\beta)-a(\gamma+\beta)+b-b=0}\) \(\large{a(\alpha+\beta-\gamma-\beta)=0}\) \(\large{a(\alpha-\gamma)=0}\) But this is not possible : Because alpha and gamma are distinct roots and alpha is not equal to zero . So , their product can not be zero . Thus the assumption that a quadratic equation has three distinct real roots is wrong . Hence a quadratic equation can not have more than 2 roots.
mathslover:
Tutorial Drill is a part of my (mathslover's) tutorials that he contributed on OpenStudy. To make those available to all, I'm trying to repost them here, for the user's benefit. Thanks! Link to tutorial 1 : http://questioncove.com/study#/updates/58613bd153dfda6d06775000 [Linear Inequalities] Link to Tutorials posted on OpenStudy: http://askandexploreforallexams.blogspot.in/
random231:
This is pretty awesome. Good Job @mathslover
mathslover:
Thanks Random231!
HanAkoSolo:
not bad my friend, not bad ^.^
pooja195:
Nice work bhaiya :)
HuskyNation:
Nice job @mathslover this tutorial will actually be helpfull for me and for other people, good job explainging it all
200082741:
Tysm
VAN1LLA:
Thank you !
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