MHB Find Cubic Function & Polynomial of Degree 3

[anemone]

Find a polynomial of degree 3 with real coefficients such that each of its roots is equal to the square of one root of the polynomial $P(x)=x^3+9x^2+9x+9$.

 

[kaliprasad]

Spoiler: my Solution

Because the square of the roots shall be roots of the required polynomial so we must have $\sqrt x$ as roots of P(X)
So $(\sqrt x)^3 + 9 x + 9 (\sqrt x) + 9 = 0$
Or $(\sqrt x)(x + 9) = - 9(x+1)$
Or squaring $x(x+9)^2 = 81(x+1)^2$
Or $x(x^2 + 18 x + 81) = 81 x^2 + 162x + 81$
Or $x^3 - 63x^2 - 81 x - 81 = 0$
This is the required equation

 

[solakis1]

Find a polynomial of degree 3 with real coefficients such that each of its roots is equal to the square of one root of the polynomial $P(x)=x^3+9x^2+9x+9$.

Does the above mean that if a,b,c are the 3 roots of the above polynomial then the the roots of the new polynomial must be $a^2 ,b^2 , c^2$ ??

 

[solakis1]

Spoiler: my Solution

Because the square of the roots shall be roots of the required polynomial so we must have $\sqrt x$ as roots of P(X)
So $(\sqrt x)^3 + 9 x + 9 (\sqrt x) + 9 = 0$
Or $(\sqrt x)(x + 9) = - 9(x+1)$
Or squaring $x(x+9)^2 = 81(x+1)^2$
Or $x(x^2 + 18 x + 81) = 81 x^2 + 162x + 81$
Or $x^3 - 63x^2 - 81 x - 81 = 0$
This is the required equation

[sp]kaliprasand must we not have also $ -\sqrt x$ .Anyway i did this problem in a different way and i got the same answer [/sp]

 

[kaliprasad]

[sp]kaliprasand must we not have also $ -\sqrt x$ .Anyway i did this problem in a different way and i got the same answer [/sp]

Spoiler: my reply

you are right and it is my mistake

as $\sqrt{x}$ is positive by definition so my line should be $\pm \sqrt{x}$ and the we get the same result because of squaring

 

[solakis1]

[sp]
Let a,b,c be the roots of P(x) = $x^3+9x^2+9x+9$
And $a^2, b^2 ,c^2 $ the roots of P(y)= $y^3+Ay^2+By+C$
From the polynomial properties we have:
P(x) = $x^3+9x^2+9x+9$=(x-a)(x-b)(x-c) $\Rightarrow$...
.a+b+c=-9...... (1)
ab+ac+bc=9 ...(2)
abc=-9 ......(3)
And also P(y)= $y^3+Ay^2+By+C$=$(y-a^2)(y-b^2)(y-c^2)\Rightarrow $
$a^2+b^2+c^2=-A$..........(4)
$a^2b^2 +a^2c^2+b^2c^2=B$......(5)
$ a^2b^2c^2=-C$............(6)
From(3) we have:$a^2b^2c^2=81$ hence
C=-81.................(7)
From (1) and squaring we get :
$(a+b+c)^2=81\Rightarrow a^2+b^2+c^2= 81-2(ab+ac+bc) $ and using (2) we get :
$a^2+b^2+c^2= 63$ and hence :
A= -63...............(8)
From (2) and squaring we get:
$(ab+ac+bc)^2=81\Rightarrow a^2b^2+a^2c^2+b^2c^2+2(a^2bc+ab^2c+abc^2)=81\Rightarrow
a^2b^2+a^2c^2+b^2c^2+2(a(abc)+b(abc)+c(abc))=81$ and using (3) we have:
$a^2b^2+a^2c^2+b^2c^2 =81-2(-9a-9b-9c)=81+18(a+b+c)$ and using (1) we have:
$a^2b^2+a^2c^2+b^2c^2=81+18(-9)=-81$ hence:
B=-81...............(9)
And P(y)= $y^3+Ay^2+By+C =y^3-63y^2-81y-81$ by using (7),(8),(9)
[/sp]