Question Number 35336 by jasno91 last updated on 18/May/18
$$\sqrt{\mathrm{2}{x}^{\mathrm{2}} }+\mathrm{7}{x}+\mathrm{5}\sqrt{\mathrm{2}}=\mathrm{0} \\ $$
Commented by Rasheed.Sindhi last updated on 18/May/18
$$\mathrm{Is}\:\mathrm{it}\:\sqrt{\mathrm{2}}\:{x}^{\mathrm{2}} \:? \\ $$
Answered by ajfour last updated on 18/May/18
$$\sqrt{\mathrm{2}}{x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{5}\sqrt{\mathrm{2}}=\mathrm{0} \\ $$$$\sqrt{\mathrm{2}}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{2}{x}+\mathrm{5}\sqrt{\mathrm{2}}=\mathrm{0} \\ $$$${x}\left(\sqrt{\mathrm{2}}{x}+\mathrm{5}\right)+\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{2}}{x}+\mathrm{5}\right)=\mathrm{0} \\ $$$$\left(\sqrt{\mathrm{2}}{x}+\mathrm{5}\right)\left({x}+\sqrt{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\:{x}=\frac{−\mathrm{5}}{\:\sqrt{\mathrm{2}}}\:,\:−\sqrt{\mathrm{2}}\:. \\ $$
Answered by Rio Mike last updated on 18/May/18
$${x}=\:\frac{−\mathrm{7}\pm\sqrt{\mathrm{7}^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{2}}\:×\:\mathrm{5}\sqrt{\mathrm{2}}}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$${x}=\:\frac{−\mathrm{7}\pm\sqrt{\mathrm{49}−\mathrm{4}×\mathrm{2}×\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$${x}=\:\frac{−\mathrm{7}\pm\sqrt{\mathrm{9}}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$${x}=\:\frac{−\mathrm{7}\pm\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$${x}=\:\frac{−\mathrm{7}\pm\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$${Either}\:{x}=\:\frac{−\mathrm{7}+\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{4}}\:\:{or}\:{x}=\:\frac{−\mathrm{7}−\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$
Commented by ajfour last updated on 18/May/18
$${second}\:{last}\:{line}\:{should}\:{be} \\ $$$${x}=\frac{−\mathrm{7}\sqrt{\mathrm{2}}\pm\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{4}}\: \\ $$$${Either}\:{x}=\frac{−\mathrm{5}}{\:\sqrt{\mathrm{2}}}\:\:{or}\:{x}=\:−\sqrt{\mathrm{2}}\:. \\ $$