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Question Number 49462 by Pk1167156@gmail.com last updated on 07/Dec/18

The number of roots of the equation 2∣x∣^2 − 7∣x∣ + 6=0.

$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{2}\mid{x}\mid^{\mathrm{2}} −\:\mathrm{7}\mid{x}\mid\:+\:\mathrm{6}=\mathrm{0}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18

∣x∣=x when x>0 =−x when x<0 now consider x>0 2x^2 −7x+6=0 2x^2 −4x−3x+6=0 2x(x−2)−3(x−2)=0 (x−2)(2x−3)=0 x=2 and (3/2) when x<0 2x^2 +7x+6=0 2x^2 +4x+3x+6=0 2x(x+2)+3(x+2)=0 (x+2)(2x+3)=0 x=−2 and ((−3)/2) so x=±2 and ±(3/2)

$$\mid{x}\mid={x}\:\:{when}\:{x}>\mathrm{0} \\ $$$$\:\:\:\:\:=−{x}\:\:{when}\:{x}<\mathrm{0} \\ $$$${now}\:{consider}\:{x}>\mathrm{0} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{6}=\mathrm{0} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{3}{x}+\mathrm{6}=\mathrm{0} \\ $$$$\mathrm{2}{x}\left({x}−\mathrm{2}\right)−\mathrm{3}\left({x}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\left({x}−\mathrm{2}\right)\left(\mathrm{2}{x}−\mathrm{3}\right)=\mathrm{0} \\ $$$${x}=\mathrm{2}\:\:{and}\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${when}\:{x}<\mathrm{0} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{6}=\mathrm{0} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}{x}+\mathrm{6}=\mathrm{0} \\ $$$$\mathrm{2}{x}\left({x}+\mathrm{2}\right)+\mathrm{3}\left({x}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\left({x}+\mathrm{2}\right)\left(\mathrm{2}{x}+\mathrm{3}\right)=\mathrm{0} \\ $$$${x}=−\mathrm{2}\:{and}\:\frac{−\mathrm{3}}{\mathrm{2}} \\ $$$${so}\:{x}=\pm\mathrm{2}\:\:{and}\:\pm\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Answered by mr W last updated on 07/Dec/18

let t=∣x∣≥0 ⇒2t^2 −7t+6=0 ⇒(2t−3)(t−2)=0 ⇒t=(3/2), 2 (both >0⇒ok) ⇒x=±(3/2), ±2

$${let}\:{t}=\mid{x}\mid\geqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{t}^{\mathrm{2}} −\mathrm{7}{t}+\mathrm{6}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{2}{t}−\mathrm{3}\right)\left({t}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow{t}=\frac{\mathrm{3}}{\mathrm{2}},\:\mathrm{2}\:\left({both}\:>\mathrm{0}\Rightarrow{ok}\right) \\ $$$$\Rightarrow{x}=\pm\frac{\mathrm{3}}{\mathrm{2}},\:\pm\mathrm{2} \\ $$

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