Question Number 82644 by Power last updated on 23/Feb/20
Commented by Power last updated on 23/Feb/20
$$\mathrm{sorry}\:\:\mathrm{it}\:\mathrm{worked} \\ $$
Commented by john santu last updated on 23/Feb/20
$$\sqrt{{x}}\:=\:{t}\:\Rightarrow\:{t}^{\mathrm{4}} \:+\:\mathrm{12}{t}−\mathrm{5}\:=\mathrm{0} \\ $$$${let}\:{x}\:+\mathrm{2}\sqrt{{x}}\:=\:{t}^{\mathrm{2}} \:+\mathrm{2}{t}\:=\:{k} \\ $$$${with}\:{Horner}\:{method}\:{we}\:{get}\: \\ $$$${t}^{\mathrm{4}} +\mathrm{12}{t}−\mathrm{5}\:=\:\left({t}^{\mathrm{2}} +\mathrm{2}{t}−{k}\right)\left({t}^{\mathrm{2}} −\mathrm{2}{t}+{k}+\mathrm{4}\right)+\:\left(\mathrm{4}−\mathrm{4}{k}\right){t}\:+{k}^{\mathrm{2}} +\mathrm{4}{k}−\mathrm{5} \\ $$
Answered by MJS last updated on 23/Feb/20
$${x}^{\mathrm{2}} +\mathrm{12}\sqrt{{x}}=\mathrm{5} \\ $$$$\mathrm{12}\sqrt{{x}}=\mathrm{5}−{x}^{\mathrm{2}} \\ $$$$\mathrm{squaring}\:\Rightarrow\:\mathrm{we}\:\mathrm{might}\:\mathrm{get}\:\mathrm{false}\:\mathrm{solutions} \\ $$$$... \\ $$$${x}^{\mathrm{4}} −\mathrm{10}{x}^{\mathrm{2}} −\mathrm{144}{x}+\mathrm{25}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{25}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{x}=\mathrm{3}\pm\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{but}\:{x}=\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{is}\:\:\mathrm{false} \\ $$$$\Rightarrow\:{x}=\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{1} \\ $$