Solve:

x^4-10x^3+42x^2-82x+65=0 the product of two of the roots of this equation is 13. The answers are complex numbers.

Thanks for helping me(Rofl)

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- May 18th 2008, 03:39 AMAnand Parikhzeros of polynomials
Solve:

x^4-10x^3+42x^2-82x+65=0 the product of two of the roots of this equation is 13. The answers are complex numbers.

Thanks for helping me(Rofl) - May 18th 2008, 04:22 AMmr fantastic
- May 18th 2008, 06:40 AMCaptainBlack
Because this is a quartic with real coefficients this factorises into a pair of real quadratics:

$\displaystyle x^4-10x^3+42x^2-82x+65=(x^2+a x +b)(x^2+ c x +d)$

where $\displaystyle a, b, c, d $ are all real. Also as the product of a pair of roots is $\displaystyle 13$, this suggests we try $\displaystyle b=13$ and $\displaystyle d=5$. Doing this we find:

$\displaystyle x^4-10x^3+42x^2-82x+65=(x^2-6x+13)(x^2-4x+5)$

and from there it is easy.

RonL - May 18th 2008, 08:01 AMSoroban
Hello, Anand Parikh!

Quote:

Solve: .$\displaystyle f(x) \:=\:x^4-10x^3+42x^2-82x+65\:=\:0$

The product of two of the roots of this equation is 13.

The roots are complex numbers.

lucky . . .*very*

Since the product of two roots is 13,

. . I**guessed**that the roots were: .$\displaystyle 2 \pm 3i\,\text{ or }\,3\pm2i$

If the roots are: .$\displaystyle 2 \pm3i$

. . then: .$\displaystyle \left(x - [2 + 3i]\right)\,\text{ and }\,\left(x-[2-3i]\right)$ .are factors of $\displaystyle f(x).$

Hence: .$\displaystyle (x-2-3i)(x-2+3i) \:=\:x^2-4x+13$ .is a factor of $\displaystyle f(x).$

. . But $\displaystyle x^2-4x+13$ does**not**divide $\displaystyle f(x).$

If the roots are: .$\displaystyle 3 \pm2i$

. . then: .$\displaystyle \left(x - [3+2i]\right),\text{ and }\, \left(x - [3-2i]\right)$ .are factors of $\displaystyle f(x).$

Hence: .$\displaystyle (x-3-2i)(x-3+2i) \:=\:x^2 -6x + 13$ .is a factor of $\displaystyle f(x).$

. . And we have: .$\displaystyle f(x)\;=\;(x^2-6x+13)(x^2-4x+5) \quad\hdots\quad \text{ta-}DAA!$