Question Number 26262 by soyebshaikh41@gmail.com last updated on 23/Dec/17
$$\left(\mathrm{x}+\mathrm{5}\right)\:\left(\mathrm{x}−\mathrm{2}\right)=\mathrm{17}−\mathrm{x}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x} \\ $$
Answered by Rasheed.Sindhi last updated on 23/Dec/17
$$\mathrm{x}^{\mathrm{2}} +\mathrm{3x}−\mathrm{10}−\mathrm{17}+\mathrm{x}=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{4x}−\mathrm{27}=\mathrm{0} \\ $$$$\mathrm{x}=\frac{−\left(\mathrm{4}\right)\pm\sqrt{\left(\mathrm{4}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(−\mathrm{27}\right)}}{\mathrm{2}\left(\mathrm{1}\right)} \\ $$$$\:\:\:\:\:=\frac{−\mathrm{4}\pm\sqrt{\mathrm{16}+\mathrm{108}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:=\frac{−\mathrm{4}\pm\sqrt{\mathrm{124}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:=\frac{−\mathrm{4}\pm\mathrm{2}\sqrt{\mathrm{31}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:=−\mathrm{2}\pm\sqrt{\mathrm{31}} \\ $$
Commented by prakash jain last updated on 25/Dec/17
Hi Rasheed, can you look at Q26312
Commented by Rasheed.Sindhi last updated on 26/Dec/17
$$\mathcal{TH}\alpha{nks}\:\mathcal{S}{ir},\:{once}\:{I}\:{tried}\:{but} \\ $$$${couldn}'{t}\:{solve}.\:{Now}\:{I}'{ll}\:{try}\:{again}. \\ $$