# 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.

Solution 2:

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

Solution 3:

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

Solution:

Let y=x^{2} and x=√y

By substitution;

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

Solution:

Let y=1/x and x=1/y

By substitution;

, multiply y^{2} to equation

Interchange y by x;

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### Dan

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