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Question-174303




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

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