Problem in finding quad. eqn. from the roots

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To find the quadratic equation with roots α+2 and ß+2 from the original equation ax^2 + bx + c = 0, one can use the sum and product of roots to derive the new equation as ax^2 - (4a-b)x + (4a-2b+c) = 0. An alternative method involves substituting x with (x-2) in the original equation, which yields the same result. This substitution method is not commonly mentioned in resources, leading to confusion. Understanding that both approaches lead to the same equation clarifies the process. The discussion highlights the validity of both methods for deriving the new quadratic equation.
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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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