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Question Number 35336 by jasno91 last updated on 18/May/18

(√(2x^2 ))+7x+5(√2)=0

$$\sqrt{\mathrm{2}{x}^{\mathrm{2}} }+\mathrm{7}{x}+\mathrm{5}\sqrt{\mathrm{2}}=\mathrm{0} \\ $$

Commented by Rasheed.Sindhi last updated on 18/May/18

Is it (√2) x^2 ?

$$\mathrm{Is}\:\mathrm{it}\:\sqrt{\mathrm{2}}\:{x}^{\mathrm{2}} \:? \\ $$

Answered by ajfour last updated on 18/May/18

(√2)x^2 +7x+5(√2)=0 (√2)x^2 +5x+2x+5(√2)=0 x((√2)x+5)+(√2)((√2)x+5)=0 ((√2)x+5)(x+(√2))=0 ⇒ x=((−5)/(√2)) , −(√2) .

$$\sqrt{\mathrm{2}}{x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{5}\sqrt{\mathrm{2}}=\mathrm{0} \\ $$$$\sqrt{\mathrm{2}}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{2}{x}+\mathrm{5}\sqrt{\mathrm{2}}=\mathrm{0} \\ $$$${x}\left(\sqrt{\mathrm{2}}{x}+\mathrm{5}\right)+\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{2}}{x}+\mathrm{5}\right)=\mathrm{0} \\ $$$$\left(\sqrt{\mathrm{2}}{x}+\mathrm{5}\right)\left({x}+\sqrt{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\:{x}=\frac{−\mathrm{5}}{\sqrt{\mathrm{2}}}\:,\:−\sqrt{\mathrm{2}}\:. \\ $$

Answered by Rio Mike last updated on 18/May/18

x= ((−7±(√(7^2 −4(√2) × 5(√2))))/(2(√2))) x= ((−7±(√(49−4×2×5)))/(2(√2))) x= ((−7±(√9))/(2(√2))) x= ((−7±3)/(2(√2))) x= ((−7±3(√2))/4) Either x= ((−7+3(√2))/4) or x= ((−7−3(√2))/4)

$${x}=\:\frac{−\mathrm{7}\pm\sqrt{\mathrm{7}^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{2}}\:×\:\mathrm{5}\sqrt{\mathrm{2}}}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$${x}=\:\frac{−\mathrm{7}\pm\sqrt{\mathrm{49}−\mathrm{4}×\mathrm{2}×\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$${x}=\:\frac{−\mathrm{7}\pm\sqrt{\mathrm{9}}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$${x}=\:\frac{−\mathrm{7}\pm\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$${x}=\:\frac{−\mathrm{7}\pm\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$${Either}\:{x}=\:\frac{−\mathrm{7}+\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{4}}\:\:{or}\:{x}=\:\frac{−\mathrm{7}−\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$

Commented by ajfour last updated on 18/May/18

second last line should be x=((−7(√2)±3(√2))/4) Either x=((−5)/(√2)) or x= −(√2) .

$${second}\:{last}\:{line}\:{should}\:{be} \\ $$$${x}=\frac{−\mathrm{7}\sqrt{\mathrm{2}}\pm\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{4}}\: \\ $$$${Either}\:{x}=\frac{−\mathrm{5}}{\sqrt{\mathrm{2}}}\:\:{or}\:{x}=\:−\sqrt{\mathrm{2}}\:. \\ $$

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