Question Number 174303 by dragan91 last updated on 29/Jul/22
Answered by a.lgnaoui last updated on 29/Jul/22
$${x}^{\mathrm{2}} −\mathrm{10}{x}−{ax}+\mathrm{10}{a}+\mathrm{1}={x}^{\mathrm{2}} +{cx}+{bx}+{bc} \\ $$$${x}^{\mathrm{2}} −\left(\mathrm{10}+{a}\right){x}+\mathrm{10}{a}+\mathrm{1}={x}^{\mathrm{2}} +\left({b}+{c}\right){x}+{bc} \\ $$$$−\left(\mathrm{10}+{a}\right)={b}+{c} \\ $$$$\mathrm{10}{a}+\mathrm{1}\:\:\:\:\:\:\:={bc} \\ $$$${b},{c}\:{verifie}\:\:{l}\:{equation}: \\ $$$${z}^{\mathrm{2}} −\left(\mathrm{10}+{a}\right){z}+\mathrm{10}{a}+\mathrm{1}=\mathrm{0} \\ $$$$\Delta=\left(\mathrm{10}+{a}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{10}{a}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:={a}^{\mathrm{2}} −\mathrm{20}{a}+\mathrm{96}=\left({a}−\mathrm{12}\right)\left({a}−\mathrm{8}\right) \\ $$$$\bullet{a}=\mathrm{8} \\ $$$$\:\:\:\:\:{b}+{c}\:\:\:=−\mathrm{18}\:\:\:{bc}=\mathrm{81} \\ $$$${z}^{\mathrm{2}} +\mathrm{18}{z}+\mathrm{81}=\left({z}+\mathrm{9}\right)^{\mathrm{2}} \:\:\:\:\:\: \\ $$$${donc}\:\:\:\:\:{a}=\mathrm{8};{b}={c}=−\mathrm{9} \\ $$$$\bullet{a}=\mathrm{12}\:\:\:\:\:\: \\ $$$${b}+{c}=−\mathrm{22}\:\:\:\:\:\:\:\:\:\:{bc}=\mathrm{121}\: \\ $$$${z}^{\mathrm{2}} +\mathrm{22}{z}+\mathrm{121}=\left({z}+\mathrm{11}\right)^{\mathrm{2}} \:\:\:\:\:\:\:\: \\ $$$$\:\:\:{donc}\:\:\:\:{a}=\mathrm{12};\:\:{b}={c}=−\mathrm{11}\: \\ $$
Commented by Tawa11 last updated on 29/Jul/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$