find the cubic polynomial

Oct 2009
56
0
Let x₁, x₂, x₃ be the roots of f(x)=x^3+7x^2-8x+3.
Find the cubic polynomial whose roots are
a) x₁^2, x₂^2, x₃^2
b) 1/(x₁), 1/(x₂), 1/(x₃)
Because we are learning resolvent cubic now, so I think that this may relate to it. Can someone tell me how to find them?
 

simplependulum

MHF Hall of Honor
Jan 2009
715
427
Let x₁, x₂, x₃ be the roots of f(x)=x^3+7x^2-8x+3.
Find the cubic polynomial whose roots are
a) x₁^2, x₂^2, x₃^2
b) 1/(x₁), 1/(x₂), 1/(x₃)
Because we are learning resolvent cubic now, so I think that this may relate to it. Can someone tell me how to find them?
I tell you the technique

Let \(\displaystyle y = x_1^2 \) as a root of the new polynomial we are looking for . The key is to write the original root as the subject . ie. \(\displaystyle x_1 = \sqrt{y} \)

Then sub. back to the equation given :

\(\displaystyle \sqrt{y}^3 + 7y - 8\sqrt{y} + 3 = 0 \)

\(\displaystyle \sqrt{y} ( y-8) = -(7y+3) \)

Squaring gives :

\(\displaystyle y(y-8)^2 = (7y+3)^2 \) I leave the expansion to you .

Try to do the second one by this method .
 
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Mar 2010
715
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Let x₁, x₂, x₃ be the roots of f(x)=x^3+7x^2-8x+3. Find the cubic polynomial whose roots are x₁^2, x₂^2, x₃^2
Using Vieta's relations, we see that \(\displaystyle x_{1}+x_{2}+x_{3} = -7\), \(\displaystyle x_{1}x_{2}+x_{1}x_{3}+x_{3}x_{2} = -8[/Math], and \(\displaystyle x_{1}x_{2}x_{3} = -3\). So \(\displaystyle x_{1}^{2}+x_{2}^{2}+x_{3}^2= (x_{1}+x_{2}+x_{3})^2-2\left(x_{1}x_{2}+x_{1}x_{3}+x_{3}x_{2}\right) = (-7)^2-2(-8) = 65
\)\)\(\displaystyle and \(\displaystyle x_{1}^2x_{2}^2+x_{1}^2x_{3}^2+x_{3}^2x_{2}^2 = (x_{1}x_{2}+x_{1}x_{3}+x_{3}x_{2})^2-2(x_{1}x_{2}x_{3})(x_{1}+x_{2}+x_{3}) \) \(\displaystyle = (-8)^2-2(-3)(-7) = 22 \) and \(\displaystyle x_{1}^2x_{2}^2x_{3}^2 = (x_{1}x_{2}x_{3})^2 = (-3)^2 = 9 \), which gives \(\displaystyle x^3+65x^2+22x+9\). The signs alternate, so the required equation is \(\displaystyle x^3-65 x^2+22 x-9\), which simplependulum's equality yields as well.\)
 
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