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Solve-in-R-15-x-2-3x-4-7-x-2-7x-10-x-2-4x-21-1-0-




Question Number 220519 by hardmath last updated on 14/May/25
Solve in   R  ((15)/(x^2  - 3x + 4))  +  (7/(x^2  + 7x))  +  ((10)/(x^2  + 4x - 21))  + 1 = 0
$$\mathrm{Solve}\:\mathrm{in}\:\:\:\mathbb{R} \\ $$$$\frac{\mathrm{15}}{\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{3x}\:+\:\mathrm{4}}\:\:+\:\:\frac{\mathrm{7}}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{7x}}\:\:+\:\:\frac{\mathrm{10}}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{4x}\:-\:\mathrm{21}}\:\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$
Commented by Nicholas666 last updated on 14/May/25
      we can only solve it by Newton Raphson method,    then we get a real solution for x     x_1  = −5.02905    x_2  = −0.29198    x_3  = 2.81703   that′s the approach I found,       But can you show me how  you do it?
$$ \\ $$$$\:\:\:\:\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{by}\:\mathrm{Newton}\:\mathrm{Raphson}\:\mathrm{method}, \\ $$$$\:\:\mathrm{then}\:\mathrm{we}\:\mathrm{get}\:\mathrm{a}\:\mathrm{real}\:\mathrm{solution}\:\mathrm{for}\:{x} \\ $$$$\:\:\:{x}_{\mathrm{1}} \:=\:−\mathrm{5}.\mathrm{02905} \\ $$$$\:\:{x}_{\mathrm{2}} \:=\:−\mathrm{0}.\mathrm{29198} \\ $$$$\:\:{x}_{\mathrm{3}} \:=\:\mathrm{2}.\mathrm{81703} \\ $$$$\:\mathrm{that}'\mathrm{s}\:\mathrm{the}\:\mathrm{approach}\:\mathrm{I}\:\mathrm{found},\: \\ $$$$\:\:\:\:\mathrm{But}\:\mathrm{can}\:\mathrm{you}\:\mathrm{show}\:\mathrm{me}\:\mathrm{how}\:\:\mathrm{you}\:\mathrm{do}\:\mathrm{it}?\:\:\:\: \\ $$$$ \\ $$

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