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Question Number 11070 by ABD last updated on 10/Mar/17

m∈Z , x^2 −(m+1)x+m^2 −16=0 f(x_1 )=f(x_2 )=0 ,x_1 <0, 0<x_2 ⇒Σm=?

$${m}\in{Z}\:,\:{x}^{\mathrm{2}} −\left({m}+\mathrm{1}\right){x}+{m}^{\mathrm{2}} −\mathrm{16}=\mathrm{0} \\ $$ $${f}\left({x}_{\mathrm{1}} \right)={f}\left({x}_{\mathrm{2}} \right)=\mathrm{0}\:\:,{x}_{\mathrm{1}} <\mathrm{0},\:\mathrm{0}<{x}_{\mathrm{2}} \:\Rightarrow\Sigma{m}=? \\ $$

Answered by mrW1 last updated on 10/Mar/17

x=(((m+1)±(√((m+1)^2 −4(m^2 −16))))/2) x=(((m+1)±(√(m^2 +2m+1−4m^2 +64)))/2) x=(((1+m)±(√(65+2m−3m^2 )))/2) x=(((1+m)±(√(65+(1/3)−3(m^2 −2×(1/3)m+(1/9)))))/2) x=(((1+m)±(√(((196)/3)−3(m−(1/3))^2 )))/2) since x_1 and x_2 should be different, ((196)/3)−3(m−(1/3))^2 >0 (m−(1/3))^2 <((196)/9) ∣m−(1/3)∣<(√((196)/9))=((14)/3) −((14)/3)<m−(1/3)<((14)/3) m>−((14)/3)+(1/3)=−((13)/3) since m ∈Z, ⇒m≥−4 m<((14)/3)+(1/3)=((15)/3)=5 ⇒−4≤m≤4 ⇒Σm=0

$${x}=\frac{\left({m}+\mathrm{1}\right)\pm\sqrt{\left({m}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\left({m}^{\mathrm{2}} −\mathrm{16}\right)}}{\mathrm{2}} \\ $$ $${x}=\frac{\left({m}+\mathrm{1}\right)\pm\sqrt{{m}^{\mathrm{2}} +\mathrm{2}{m}+\mathrm{1}−\mathrm{4}{m}^{\mathrm{2}} +\mathrm{64}}}{\mathrm{2}} \\ $$ $${x}=\frac{\left(\mathrm{1}+{m}\right)\pm\sqrt{\mathrm{65}+\mathrm{2}{m}−\mathrm{3}{m}^{\mathrm{2}} }}{\mathrm{2}} \\ $$ $${x}=\frac{\left(\mathrm{1}+{m}\right)\pm\sqrt{\mathrm{65}+\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{3}\left({m}^{\mathrm{2}} −\mathrm{2}×\frac{\mathrm{1}}{\mathrm{3}}{m}+\frac{\mathrm{1}}{\mathrm{9}}\right)}}{\mathrm{2}} \\ $$ $${x}=\frac{\left(\mathrm{1}+{m}\right)\pm\sqrt{\frac{\mathrm{196}}{\mathrm{3}}−\mathrm{3}\left({m}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} }}{\mathrm{2}} \\ $$ $${since}\:{x}_{\mathrm{1}} \:{and}\:{x}_{\mathrm{2}} \:{should}\:{be}\:{different}, \\ $$ $$\frac{\mathrm{196}}{\mathrm{3}}−\mathrm{3}\left({m}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} >\mathrm{0} \\ $$ $$\left({m}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} <\frac{\mathrm{196}}{\mathrm{9}} \\ $$ $$\mid{m}−\frac{\mathrm{1}}{\mathrm{3}}\mid<\sqrt{\frac{\mathrm{196}}{\mathrm{9}}}=\frac{\mathrm{14}}{\mathrm{3}} \\ $$ $$−\frac{\mathrm{14}}{\mathrm{3}}<{m}−\frac{\mathrm{1}}{\mathrm{3}}<\frac{\mathrm{14}}{\mathrm{3}} \\ $$ $${m}>−\frac{\mathrm{14}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}=−\frac{\mathrm{13}}{\mathrm{3}} \\ $$ $${since}\:{m}\:\in\mathbb{Z}, \\ $$ $$\Rightarrow{m}\geqslant−\mathrm{4} \\ $$ $${m}<\frac{\mathrm{14}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{15}}{\mathrm{3}}=\mathrm{5} \\ $$ $$\Rightarrow−\mathrm{4}\leqslant{m}\leqslant\mathrm{4} \\ $$ $$ \\ $$ $$\Rightarrow\Sigma{m}=\mathrm{0} \\ $$

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