Question Number 224095 by fitraha last updated on 19/Aug/25
$${how}\:{to}\:{prove}\:{that}\:\:{x}\:+\:\mathrm{9}\:=\:{x}\:{is}\:{not}\:{has}\:{solution} \\ $$$${because}\: \\ $$$$\left({x}\:+\:\mathrm{9}\right)^{\mathrm{2}} \:=\:{x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} \:+\:\mathrm{18}{x}\:+\:\mathrm{81}\:=\:{x}^{\mathrm{2}} \\ $$$$\mathrm{18}{x}\:=\:−\mathrm{81} \\ $$$${x}\:=\:−\:\frac{\mathrm{81}}{\mathrm{18}}\:=\:−\frac{\mathrm{9}}{\mathrm{2}} \\ $$
Commented by mr W last updated on 19/Aug/25
$$\left(−\mathrm{5}\right)^{\mathrm{2}} =\left(\mathrm{5}\right)^{\mathrm{2}} \:{doesn}'{t}\:{mean}\:−\mathrm{5}=\mathrm{5}. \\ $$
Commented by Frix last updated on 19/Aug/25
$$\mathrm{Squaring}\:\mathrm{introduces}\:\mathrm{false}\:\mathrm{solutions}. \\ $$$${x}=\mathrm{2} \\ $$$${x}+\mathrm{5}=\mathrm{7} \\ $$$$\left({x}+\mathrm{5}\right)^{\mathrm{2}} =\mathrm{49} \\ $$$${x}^{\mathrm{2}} +\mathrm{10}{x}−\mathrm{24}=\mathrm{0} \\ $$$${x}=−\mathrm{12}\vee{x}=\mathrm{2} \\ $$
Answered by RedstoneGG4 last updated on 19/Aug/25
$$\mathrm{What}\:\mathrm{are}\:\mathrm{you}\:\mathrm{doing}? \\ $$$$\cancel{{x}}\:+\:\mathrm{9}\:=\:\cancel{{x}}\:\rightarrow\:\mathrm{9}\:=\:\mathrm{0}\:× \\ $$