Question Number 177648 by BaliramKumar last updated on 07/Oct/22 | ||
$${If}\:\:\frac{{a}\:+\:{b}}{\:\sqrt{{ab}}}\:=\:\frac{\mathrm{4}}{\mathrm{1}}\:{then}\:\:\:\:{a}\::\:{b}\:=\:?\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\:{a}>{b}\:\right] \\ $$ | ||
Answered by Rasheed.Sindhi last updated on 07/Oct/22 | ||
$${a}+{b}=\mathrm{4}\sqrt{{ab}}\: \\ $$ $${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{16}{ab}−\mathrm{2}{ab}=\mathrm{14}{ab} \\ $$ $$\frac{{a}\:}{{b}}+\frac{{b}}{{a}}=\mathrm{14} \\ $$ $${x}+\frac{\mathrm{1}}{{x}}=\mathrm{14};\:\left[\:\frac{{a}}{{b}}={x}\right] \\ $$ $${x}^{\mathrm{2}} −\mathrm{14}{x}+\mathrm{1}=\mathrm{0} \\ $$ $${x}=\frac{\mathrm{14}\pm\sqrt{\mathrm{196}−\mathrm{4}}}{\mathrm{2}}=\frac{\mathrm{14}\pm\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{7}\pm\mathrm{4}\sqrt{\mathrm{3}} \\ $$ $$\frac{{a}}{{b}}=\frac{\mathrm{7}\pm\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{1}} \\ $$ $$\because\:{a}>{b} \\ $$ $$\therefore\:\:{a}:{b}=\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\:\right):\mathrm{1} \\ $$ | ||
Commented byBaliramKumar last updated on 07/Oct/22 | ||
$${nice}\:{solution} \\ $$ | ||