root equation .

[ringokid]

Need help with this I dont know how to do it.

Show that the equation 8x^5 + 77x^4 + 2x^3 + 8x^2 - 93x + 42 = 0 has exactly 3 real roots.

Thanks,

 

[BigGlenntheHeavy]

$\displaystyle Descartes' \ Rule \ of \ Signs \ gives:$

$\displaystyle f(x) \ = \ 8x^5+77x^4+2x^3+8x^2-93x+42 \ \implies \ 2 \ changes$

$\displaystyle f(-x) \ = \ -8x^5+77x^4-2x^3+8x^2+93x+42 \ \implies \ 3 \ changes$

$\displaystyle Hence, \ we \ have \ 4 \ possibilities, \ to \ wit:$

\(\displaystyle 2^+,3^-, 0^I \ or \ \ 2^+,1^-,2^I \ or \ \ 0^+,3^-,2^I, \ or \ 0^+,1^-,4^I \\)

$\displaystyle Now, \ if \ we \ assume \ that \ the \ equations \ has \ 3 \ real \ roots, \ then,$

$\displaystyle only \ 2 \ possibilities \ exist, \ viz., \ 2^+,1^-,2^I \ or \ 0^+,3^-.2^I$

$\displaystyle Now, \ taking \ out \ my \ trusty \ TI-89, \ I \ get \ the \ following \ roots:$

$\displaystyle (.62347538,0),(.62348262,0), \ and \ (-9.623475,0) \ which \ coincides \ with \ 2^+,1^-,2^I$

$\displaystyle Note; \ You \ could \ also \ find \ the \ roots \ by \ the \ method \ of \ exhaustion \ or \ Newton's$

$\displaystyle Method \ if \ you \ prefer \ a \ lot \ of \ tedious \ grunt \ work.$

 

[galactus]

If we factor:

$\displaystyle (x^{2}+9x-6)(8x^{3}+5x^{2}+5x-7)$

The quadratic part discriminant, $\displaystyle b^{2}-4ac=81-4(1)(-6)=105$, is > 0. Therefore, it has two real roots.

2 down.

The cubic part discriminant, $\displaystyle b^{2}c^{2}-4c^{3}-4b^{3}d-27d^{2}+18bcd=(5)^{2}(5)^{2}-4(5)^{3}-4(5)(-7)-27(-7)^{2}+18(5)(5)(-7)=-4208$

Since the discriminant < 0, then it has 1 real and two complex roots.

Thus, there is 1 real from the quadratic and two real from the cubic...making three real roots. The other two are complex.

 

[ringokid]

Thx for your help.