# Roots of cubic equation

• October 10th 2009, 05:25 PM
Hellbent
Roots of cubic equation
Hi,

Kindly help me with this roots question...please

If the roots of the equation 4x³ + 7x² - 5x - 1 = 0 are α, β and γ, find the equation whose roots are
i. α + 1, β + 1 and γ + 1

ii. α², β², γ²
• October 11th 2009, 01:05 AM
red_dog
We have $x_1=\alpha, \ x_2=\beta, \ x_3=\gamma$. Then

$x_1+x_2+x_3=-\frac{7}{4}$

$x_1x_2+x_1x_3+x_2x_3=-\frac{5}{4}$

$x_1x_2x_3=\frac{1}{4}$

1) $y_1=x_1+1, \ y_2=x_2+1, \ y_3=x_3+1$

In this case the easiest way is to let $y=x+1$. Then $x=y-1$ and replace x in the equation:

$4(y-1)^3+7(y-1)^2-5(y-1)-1=0$

Now continue.

2) Let $S_1=y_1+y_2+y_3=x_1^2+x_2^2+x_3^2=(x_1+x_2+x_3)^2-2(x_1x_2+x_1x_3+x_2x_3)$

$S_2=y_1y_2+y_1y_3+y_2y_3=x_1^2x_2^2+x_1^2x_3^2+x_2 ^2x_3^2=$

$=(x_1x_2+x_1x_3+x_2x_3)^2-2x_1x_2x_3(x_1+x_2+x_3)$

$S_3=y_1y_2y_3=(x_1x_2x_3)^2$

Then the equation is $y^3-S_1y^2+S_2y-S_3=0$
• October 11th 2009, 04:18 AM
Quote:

Originally Posted by red_dog
We have $x_1=\alpha, \ x_2=\beta, \ x_3=\gamma$. Then

$x_1+x_2+x_3=-\frac{7}{4}$

$x_1x_2+x_1x_3+x_2x_3=-\frac{5}{4}$

$x_1x_2x_3=\frac{1}{4}$

1) $y_1=x_1+1, \ y_2=x_2+1, \ y_3=x_3+1$

In this case the easiest way is to let $y=x+1$. Then $x=y-1$ and replace x in the equation:

$4(y-1)^3+7(y-1)^2-5(y-1)-1=0$

Now continue.

2) Let $S_1=y_1+y_2+y_3=x_1^2+x_2^2+x_3^2=(x_1+x_2+x_3)^2-2(x_1x_2+x_1x_3+x_2x_3)$

$S_2=y_1y_2+y_1y_3+y_2y_3=x_1^2x_2^2+x_1^2x_3^2+x_2 ^2x_3^2=$

$=(x_1x_2+x_1x_3+x_2x_3)^2-2x_1x_2x_3(x_1+x_2+x_3)$

$S_3=y_1y_2y_3=(x_1x_2x_3)^2$

Then the equation is $y^3-S_1y^2+S_2y-S_3=0$

I'm having trouble with similar problems. How do you get:

$x_1+x_2+x_3=-\frac{7}{4}$

??
• October 11th 2009, 05:35 AM
red_dog
Quote:

I'm having trouble with similar problems. How do you get:

$x_1+x_2+x_3=-\frac{7}{4}$

??

If $x_1, \ x_2, \ x_3$ are the roots of equation $ax^3+bx^2+cx+d=0$ then

$x_1+x_2+x_3=-\frac{b}{a}$

$x_1x_2+x_1x_3+x_2x_3=\frac{c}{a}$

$x_1x_2x_3=-\frac{d}{a}$

These are the relations of Viete.