1super. nope my method was longer so i posted here :P
p(x) = x5 + x2 +1 have roots
x1 , x2,x3,x4,x5.
g(x) = x2-2. then the value of
g(x1).g(x2).g(x3).g(x4).g(x5) - 30g(x1.x2.x3.x4.x5) is ?
-
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6 Answers
1708\hspace{-16}$Let $\mathbf{g(x)=y=x^2-2\Leftrightarrow x^2=(y+2)}$\\\\ Now Given That $\mathbf{x^5+x^2+1=0\Leftrightarrow (x^5)=-(x^2+1)}$\\\\ $\mathbf{x^{10}=x^4+1+2x^2}$\\\\ $\mathbf{(y+2)^5=(y+2)^2+1+2y+4}$\\\\ $\mathbf{y^5+10y^4+40y^3+79y^2+74y+23=0}$\\\\ So $\mathbf{y_{1}.y_{2}.y_{3}.y_{4}.y_{5}=-23}$\\\\ Where $\mathbf{y_{1}\;,y_{2}\;,y_{3}\;,y_{4}\;,y_{5}}$ are the Roots of Given Eq.\\\\ and $\mathbf{x^5+x^2+1=0}$\\\\ So $\mathbf{x_{1}.x_{2}.x_{3}.x_{4}.x_{5}=-1}$\\\\ So $\mathbf{g(x_{1}.x_{2}.x_{3}.x_{4}.x_{5})=(x_{1}.x_{2}.x_{3}.x_{4}.x_{5})^2-2}$\\\\ So $\mathbf{y_{1}.y_{2}.y_{3}.y_{4}.y_{5}-30.(1-2)=-23+30=7}$\\\\ Where $\mathbf{g(x_{i})=y_{i}\forall i=1,2,3,4,5}$
May be some other beautiful Method, So Rishab if you have so plz post here.
Thanks
341\because \ g(x) = (x-\sqrt 2)(x+\sqrt 2) and x_1x_2x_3x_4x_5=-1
the given expression equals
g(\sqrt 2) g(-\sqrt2) - 30g(-1)=7
1@hsbhat sir i guess there's a typo in your post. g(√2) = 0 ...
btw, how do you get anything of that sort directly?
341sorry, i meant p(\sqrt 2) p(-\sqrt 2) - 30 p(-1) = 7
I use p(x) = (x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)
So
(x_1-\sqrt2)(x_2-\sqrt 2))(x_3-\sqrt2)(x_4-\sqrt 2)(x_5-\sqrt 2) = -p(\sqrt2)
Now you can finish