Beating Vieta’s Formula
Find an equation with integral coefficients whose roots are twice the roots of . Did you ever encounter this type of problem? In this topic we will discuss how to treat this problem properly and I will teach you how to speed up your calculation.
Sample Problem 1:
Find an equation with integral coefficients whose roots are twice the roots of
Factoring we have and the roots of this equation are 1 and 2. Since we are looking for another quadratic equation with roots twice of the given. The new roots must be 2 and 4.
The required equation is and finally
That is easy if the given equation is factorable.
Let a and b are the roots of equation
By, Vieta’s formula;
a+b = 3 and ab=8, since we are ask to get derive an equation twice the roots of given equation , the new equation must have the following properties,
a+b=6 and ab=8. Simply multiplying by 2.
The required Equation is
Let x be the roots of the given equation and y be the roots of the new equation.
It is stated that the roots of new equation (y) is twice the roots of given equation (x) or
y = 2x. Solving for x, we have x=y/2
Now plug the value of x in the original equation.
, and since we are ask for integral coefficient we will multiply the whole equation by 4 and interchange all y’s to x which will yield
Sample Problem 2:
Find an equation with integral coefficient whose roots are the squares of the roots of
Let y=x2 and x=√y
Interchange again y by x.
this is the required equation
Sample Problem 3:
Find an equation with integral coefficient whose roots are the reciprocals of the roots of
Let y=1/x and x=1/y
, multiply y2 to equation
Interchange y by x;
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