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Question Number 214408 by Abdullahrussell last updated on 07/Dec/24

Answered by A5T last updated on 07/Dec/24

$$Letf\left(x\right)=x^4-10x^3+37x^2-60x+32$$$$f\left(1\right)=f\left(4\right)=0\Rightarrow f\left(x\right)=(x^2-5x+4)(x^2-5x+8)$$$$\Rightarrow \mathrm{\Sigma }\frac{1}{a^2-5a+10}=\frac{1}{6}+\frac{1}{6}+\frac{1}{2}+\frac{1}{2}=1\frac{1}{3}=\frac{4}{3}$$

Commented by Abdullahrussell last updated on 08/Dec/24

$$Sir,kindlydetails$$

Answered by mr W last updated on 08/Dec/24

$$say{x}^2-5x+10=t$$$$t^2={(x^2-5x+10)}^2$$$$=x^4+25x^2+100-10x^3+20x^2-100x$$$$=x^4-10x^3+37x^2-60x+32+8x^2-40x+68$$$$=8x^2-40x+68$$$$=8(x^2-5x+10)-12$$$$=8t-12$$$$t^2-8t+12=0$$$$(t-2)(t-6)=0$$$$t_{1,2}=2,6\Rightarrow \frac{1}{t_{1,2}}=\frac{1}{2},\frac{1}{6}$$$$\Rightarrow \frac{1}{x_{1,2}^2-5x_{1,2}+10}=\frac{1}{2}$$$$\Rightarrow \frac{1}{x_{3,4}^2-5x_{3,4}+10}=\frac{1}{6}$$$$i.e.$$$$\frac{1}{a^2-5a+10}+\frac{1}{b^2-5b+10}+\frac{1}{c^2-5c+10}+\frac{1}{d^2-5d+10}$$$$=\frac{1}{2}+\frac{1}{2}+\frac{1}{6}+\frac{1}{6}$$$$=\frac{4}{3}✓$$

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