# Beating Vieta’s Formula

Find an equation with integral coefficients whose roots are twice the roots of  $x^2-3x+2=0$. 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 $x^2-3x+2=0$

Solution 1:

Factoring $x^2-3x+2=0$ we have $(x-1)(x-2)$ 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 $(x-2)(x-4)=0$ and finally $x^2-6x+8=0$

That is easy if the given equation is factorable.

Solution 2:

Let a and b are the roots of equation $x^2-3x+2=0$

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 $x^2-6x+8=0$

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.

$x^2-3x+2=0$

$(\displaystyle\frac{y}{2})^2-3(\frac{y}{2})+2=0$

$\displaystyle\frac{y^2}{4}-\displaystyle\frac{3y}{2}+2=0$
, 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 $x^2-6x+8=0$

Sample Problem 2:

Find an equation with integral coefficient whose roots are the squares of the roots of $x^2-x-4=0$

Solution:

Let y=x2 and x=√y

By substitution;

$x^2-x-4=0$

$(\sqrt{y})^2-\sqrt{y}-4=0$

$y-4=\sqrt{y}$

$(y-4)^2=(\sqrt{y})^2$

$y^2-8y+16=y$

$y^2-9y+16=0$

Interchange again y by x.

$x^2-9x+16=0$ this is the required equation

Sample Problem 3:

Find an equation with integral coefficient whose roots are the reciprocals of the roots of $2x^2+5x-9=0$

Solution:

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

By substitution;

$2(\frac{1}{y})^2+5(\frac{1}{y})-9=0$

$\frac{2}{y^2}+\frac{5}{y}-9=0$ , multiply y2 to equation

$2+5y-9y^2=0$

Interchange y by x;

$2+5x-9x^2=0$

Image Credit: wikipedia

### Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.