Question Number 98907 by Ramajunan last updated on 17/Jun/20
Answered by mr W last updated on 17/Jun/20
$$\frac{\left(\mathrm{5}+{x}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{5}+{x}\right){x}}=−\frac{\mathrm{5}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} }{\mathrm{2}×\mathrm{5}{x}} \\ $$$${x}^{\mathrm{3}} +\mathrm{15}{x}^{\mathrm{2}} +\mathrm{26}{x}−\mathrm{240}=\mathrm{0} \\ $$$$\left({x}−\mathrm{3}\right)\left({x}+\mathrm{8}\right)\left({x}+\mathrm{10}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{3} \\ $$
Commented by Brokris last updated on 17/Jun/20
$$\mathrm{pls}\:\mathrm{what}\:\mathrm{formula}\:\mathrm{did}\:\mathrm{you}\:\mathrm{use}? \\ $$
Commented by mr W last updated on 17/Jun/20
$${law}\:{of}\:{cosines} \\ $$$$\mathrm{cos}\:\alpha=.... \\ $$$$\mathrm{cos}\:\beta=.... \\ $$$${a}+\beta=\mathrm{180}° \\ $$$$\mathrm{cos}\:\alpha=−\mathrm{cos}\:\beta \\ $$
Commented by mr W last updated on 17/Jun/20
