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Question Number 96072 by i jagooll last updated on 29/May/20

(x^2 +24x+24).(x^2 +x+24)= 24x^2

$$\left(\mathrm{x}^{\mathrm{2}} +\mathrm{24x}+\mathrm{24}\right).\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{24}\right)=\:\mathrm{24x}^{\mathrm{2}} \\ $$

Answered by bobhans last updated on 29/May/20

set x^2 +24 = y (y+24x)(y+x)=24x^2 y^2 +xy+24xy+24x^2 = 24x^2 y^2 +25xy = 0 y(y+25x) = 0 ⇒ { ((y=0 ⇒x = ±2i(√6))),((y+25x = 0 )) :} x^2 +25x+24 = 0 ⇒(x+1)(x+24)=0 x = −24 or −1

$$\mathrm{set}\:\mathrm{x}^{\mathrm{2}} +\mathrm{24}\:=\:\mathrm{y}\: \\ $$$$\left(\mathrm{y}+\mathrm{24x}\right)\left(\mathrm{y}+\mathrm{x}\right)=\mathrm{24x}^{\mathrm{2}} \\ $$$$\mathrm{y}^{\mathrm{2}} +\mathrm{xy}+\mathrm{24xy}+\mathrm{24x}^{\mathrm{2}} \:=\:\mathrm{24x}^{\mathrm{2}} \: \\ $$$$\mathrm{y}^{\mathrm{2}} +\mathrm{25xy}\:=\:\mathrm{0} \\ $$$$\mathrm{y}\left(\mathrm{y}+\mathrm{25x}\right)\:=\:\mathrm{0}\:\Rightarrow\:\begin{cases}{\mathrm{y}=\mathrm{0}\:\Rightarrow\mathrm{x}\:=\:\pm\mathrm{2}{i}\sqrt{\mathrm{6}}}\\{\mathrm{y}+\mathrm{25x}\:=\:\mathrm{0}\:}\end{cases} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{25x}+\mathrm{24}\:=\:\mathrm{0}\:\Rightarrow\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}+\mathrm{24}\right)=\mathrm{0} \\ $$$$\mathrm{x}\:=\:−\mathrm{24}\:\mathrm{or}\:−\mathrm{1}\: \\ $$

Commented by i jagooll last updated on 29/May/20

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