Question Number 55014 by Rio Mike last updated on 16/Feb/19
$${what}\:{is}\:{the}\:{value}\:{of}\:{t}\:{that}\:{makes}\: \\ $$$${x}^{\mathrm{2}} +\mathrm{10}{x}+{t}\:{a}\:{perfect}\:{square}? \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by peter frank last updated on 16/Feb/19
$${t}=\mathrm{25} \\ $$
Answered by Joel578 last updated on 16/Feb/19
$${x}^{\mathrm{2}} \:+\:\mathrm{10}{x}\:+\:{t}\:\equiv\:\left({x}\:+\:\sqrt{{t}}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} \:+\:\mathrm{10}{x}\:+\:{t}\:\equiv\:{x}^{\mathrm{2}} \:+\:\mathrm{2}\sqrt{{t}}{x}\:+\:{t} \\ $$$$\mathrm{10}\:=\:\mathrm{2}\sqrt{{t}}\:\rightarrow\:{t}\:=\:\mathrm{25} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 16/Feb/19
$${x}^{\mathrm{2}} +\mathrm{2}×{x}×\mathrm{5}+\mathrm{5}^{\mathrm{2}} +{t}−\mathrm{25} \\ $$$$\left({x}+\mathrm{5}\right)^{\mathrm{2}} +\left({t}−\mathrm{25}\right) \\ $$$${to}\:{make}\:{perfect}\:{square}\:\left({t}−\mathrm{25}\right)=\mathrm{0} \\ $$$${so}\:{t}=\mathrm{25} \\ $$