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Question Number 138250 by sahnaz last updated on 11/Apr/21

(x^2 −10x+6)^(x−2) >1

$$\left(\mathrm{x}^{\mathrm{2}} −\mathrm{10x}+\mathrm{6}\right)^{\mathrm{x}−\mathrm{2}} >\mathrm{1} \\ $$

Answered by bobhans last updated on 11/Apr/21

(x^2 −10x+6)^(x−2)  > (x^2 −10x+6)^0   (1)x−2=0⇒x=2  (2) x^2 −10x+6=0 ; (x−5)^2 −25+6=0  (x−5)^2 =19 ; x=5 ±(√(19))  { ((x=5+(√(19)))),((x=5−(√(19)) (reject))) :}  (3)x^2 −10x+6=1 ; x^2 −10x+5=0  (x−5)^2 −20=0 ; x=5±2(√5)  (4)x^2 −10x+6=−1  x^2 −10x+7=0  (x−5)^2 −18=0; x=5±3(√2)  { ((x=5+3(√2))),((x=5−3(√2) (reject))) :}  solution x = 2; 5+(√(19)) ; 5±2(√5) ; 5+3(√2)

$$\left({x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{6}\right)^{{x}−\mathrm{2}} \:>\:\left({x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{6}\right)^{\mathrm{0}} \\ $$ $$\left(\mathrm{1}\right){x}−\mathrm{2}=\mathrm{0}\Rightarrow{x}=\mathrm{2} \\ $$ $$\left(\mathrm{2}\right)\:{x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{6}=\mathrm{0}\:;\:\left({x}−\mathrm{5}\right)^{\mathrm{2}} −\mathrm{25}+\mathrm{6}=\mathrm{0} \\ $$ $$\left({x}−\mathrm{5}\right)^{\mathrm{2}} =\mathrm{19}\:;\:{x}=\mathrm{5}\:\pm\sqrt{\mathrm{19}}\:\begin{cases}{{x}=\mathrm{5}+\sqrt{\mathrm{19}}}\\{{x}=\mathrm{5}−\sqrt{\mathrm{19}}\:\left({reject}\right)}\end{cases} \\ $$ $$\left(\mathrm{3}\right){x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{6}=\mathrm{1}\:;\:{x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{5}=\mathrm{0} \\ $$ $$\left({x}−\mathrm{5}\right)^{\mathrm{2}} −\mathrm{20}=\mathrm{0}\:;\:{x}=\mathrm{5}\pm\mathrm{2}\sqrt{\mathrm{5}} \\ $$ $$\left(\mathrm{4}\right){x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{6}=−\mathrm{1} \\ $$ $${x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{7}=\mathrm{0} \\ $$ $$\left({x}−\mathrm{5}\right)^{\mathrm{2}} −\mathrm{18}=\mathrm{0};\:{x}=\mathrm{5}\pm\mathrm{3}\sqrt{\mathrm{2}}\:\begin{cases}{{x}=\mathrm{5}+\mathrm{3}\sqrt{\mathrm{2}}}\\{{x}=\mathrm{5}−\mathrm{3}\sqrt{\mathrm{2}}\:\left({reject}\right)}\end{cases} \\ $$ $${solution}\:{x}\:=\:\mathrm{2};\:\mathrm{5}+\sqrt{\mathrm{19}}\:;\:\mathrm{5}\pm\mathrm{2}\sqrt{\mathrm{5}}\:;\:\mathrm{5}+\mathrm{3}\sqrt{\mathrm{2}}\: \\ $$

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