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Question Number 67844 by aliesam last updated on 01/Sep/19

x^2 +∣x∣−6=0

$${x}^{\mathrm{2}} +\mid{x}\mid−\mathrm{6}=\mathrm{0} \\ $$

Commented by gunawan last updated on 01/Sep/19

case 1 x<0  x^2 −x−6=0  (x−3)(x+2)=0  x=3 ∨x=−2  solution  (2)^2 +∣−2∣−6=0  (3)^2 +∣3∣−6≠0  solution x=−2  case 2 x >0  x^2 +x−6=0  (x+3)(x−2)=0  x=−3 ∨ x=2  2^2 +∣2∣−6=0  (−3)^2 +∣−3∣−6≠0  solution is  x=2 or x=−2

$$\mathrm{case}\:\mathrm{1}\:{x}<\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{6}=\mathrm{0} \\ $$$$\left({x}−\mathrm{3}\right)\left({x}+\mathrm{2}\right)=\mathrm{0} \\ $$$${x}=\mathrm{3}\:\vee{x}=−\mathrm{2} \\ $$$$\mathrm{solution} \\ $$$$\left(\mathrm{2}\right)^{\mathrm{2}} +\mid−\mathrm{2}\mid−\mathrm{6}=\mathrm{0} \\ $$$$\left(\mathrm{3}\right)^{\mathrm{2}} +\mid\mathrm{3}\mid−\mathrm{6}\neq\mathrm{0} \\ $$$$\mathrm{solution}\:{x}=−\mathrm{2} \\ $$$$\mathrm{case}\:\mathrm{2}\:{x}\:>\mathrm{0} \\ $$$${x}^{\mathrm{2}} +{x}−\mathrm{6}=\mathrm{0} \\ $$$$\left({x}+\mathrm{3}\right)\left({x}−\mathrm{2}\right)=\mathrm{0} \\ $$$${x}=−\mathrm{3}\:\vee\:{x}=\mathrm{2} \\ $$$$\mathrm{2}^{\mathrm{2}} +\mid\mathrm{2}\mid−\mathrm{6}=\mathrm{0} \\ $$$$\left(−\mathrm{3}\right)^{\mathrm{2}} +\mid−\mathrm{3}\mid−\mathrm{6}\neq\mathrm{0} \\ $$$$\mathrm{solution}\:\mathrm{is} \\ $$$${x}=\mathrm{2}\:\mathrm{or}\:{x}=−\mathrm{2} \\ $$

Commented by mathmax by abdo last updated on 01/Sep/19

(e)⇒∣x∣^2 +∣x∣−6   let ∣x∣ =t  with t≥0 ⇒t^2 +t−6 =0  Δ =1−4(−6) =25 ⇒t_1 =((−1+5)/2) =2  and t_2 =((−1−5)/2) =−3<0(to  eliminate)  and ∣x∣ =2 ⇒x =2 or x =−2 .

$$\left({e}\right)\Rightarrow\mid{x}\mid^{\mathrm{2}} +\mid{x}\mid−\mathrm{6}\:\:\:{let}\:\mid{x}\mid\:={t}\:\:{with}\:{t}\geqslant\mathrm{0}\:\Rightarrow{t}^{\mathrm{2}} +{t}−\mathrm{6}\:=\mathrm{0} \\ $$$$\Delta\:=\mathrm{1}−\mathrm{4}\left(−\mathrm{6}\right)\:=\mathrm{25}\:\Rightarrow{t}_{\mathrm{1}} =\frac{−\mathrm{1}+\mathrm{5}}{\mathrm{2}}\:=\mathrm{2}\:\:{and}\:{t}_{\mathrm{2}} =\frac{−\mathrm{1}−\mathrm{5}}{\mathrm{2}}\:=−\mathrm{3}<\mathrm{0}\left({to}\right. \\ $$$$\left.{eliminate}\right)\:\:{and}\:\mid{x}\mid\:=\mathrm{2}\:\Rightarrow{x}\:=\mathrm{2}\:{or}\:{x}\:=−\mathrm{2}\:. \\ $$

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