Question Number 103458 by bobhans last updated on 15/Jul/20
$$\left(\mathrm{1}\right){If}\:{p}\:{and}\:{q}\:{are}\:{the}\:{roots}\:{quadratic} \\ $$$${equation}\:{x}^{\mathrm{2}} −{x}−\mathrm{4050}\:=\:\mathrm{0}\:,\:{find}\:{the}\:{value} \\ $$$${of}\:{p}^{\mathrm{2}} +{q}\:?\: \\ $$$$\left(\mathrm{2}\right)\:{if}\:{f}\left({x}\right)=\frac{\mathrm{7}{x}+{q}}{\mathrm{2}}\:{and}\:{g}\left({x}\right)=\mathrm{7}{x}+\mathrm{8}\:; \\ $$$$\left({fog}\right)\left(\mathrm{4}\right)\:=\:\mathrm{24}\:.\:{find}\:{the}\:{value}\:{of}\:{q} \\ $$
Answered by bemath last updated on 15/Jul/20
$$\left(\mathrm{1}\right)\:{x}^{\mathrm{2}} −{x}−\mathrm{4050}=\mathrm{0}\rightarrow\begin{cases}{{p}}\\{{q}}\end{cases} \\ $$$$\Leftrightarrow{p}^{\mathrm{2}} \:=\:{p}+\mathrm{4050} \\ $$$${then}\:{p}^{\mathrm{2}} +{q}\:=\:{p}+\mathrm{4050}+{q} \\ $$$$=\:{p}+{q}\:+\mathrm{4050} \\ $$$$=\:\mathrm{1}+\mathrm{4050}\:=\:\mathrm{4051}\:;\left({by}\:{Vieta}'{s}\right. \\ $$$$\left.{rule}\:\right)\bigstar \\ $$
Answered by bemath last updated on 15/Jul/20
$$\left(\mathrm{2}\right)\left({fog}\right)\left(\mathrm{4}\right)=\:{f}\left({g}\left(\mathrm{4}\right)\right) \\ $$$$\Rightarrow{g}\left(\mathrm{4}\right)=\:\mathrm{7}.\mathrm{4}\:+\:\mathrm{8}\:=\:\mathrm{36}\: \\ $$$$\Rightarrow{f}\left(\mathrm{36}\right)\:=\:\mathrm{24}\:\rightarrow\frac{\mathrm{7}.\mathrm{36}+{q}}{\mathrm{2}}\:=\:\mathrm{24} \\ $$$${q}\:=\:\mathrm{48}−\mathrm{252}\:=\:−\mathrm{204}\:\bigstar\: \\ $$