Problem in finding quad. eqn. from the roots

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The discussion focuses on deriving a quadratic equation from its roots, specifically when the roots are transformed by adding 2. Given roots α and ß of the equation ax² + bx + c = 0, the new roots α+2 and ß+2 lead to the equation ax² - (4a-b)x + (4a-2b+c) = 0. The method of substituting x with (x-2) in the original equation is confirmed to yield the same result, although this substitution technique is not widely documented.

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Sumedh
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In general, if α(alpha) and ß (beta) are roots of eqn. ax^2 +bx +c=0
then for finding the equation whose roots are α+2 and ß+2 can be done by
addition of roots (α+2+ß+2=-b/a) and product of roots (α+2)(ß+2)=c/a
By solving this we get ax^2 -(4a-b)x + (4a-2b+c)=0

The problem is this that,by replacing x in place of (x-2)in the given equation
ax^2 +bx +c=0 we get the same answer ax^2 -(4a-b)x + (4a-2b+c)=0
but this method ( replacing x by (x-2) ..) is not mentioned anywhere
i am not able to understand this method
please help .
 
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I'm not sure what you are asking. Certainly, it is true that if \alpha and \beta are roots of a quadratic equation, then the equation is of the form a(x- \alpha)(x- \beta)= 0 for some number a.

Similarly, if the roots of a quadratic equation are \alpha+ 2 and \beta+ 2, then the equation is of the form a(x- (\alpha+ 2))(x- (\beta+ 2)= 0 which is the same as a((x- 2)- \alpha)(x- 2)- \beta)= 0, the original equation with "x" replaced by "x- 2".
 
thanks i got it:smile:
 

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