Solve the following equation:
x+1x=212.x + \dfrac{1}{x} = 2\dfrac{1}{2}.x+x1=221.
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Given, x+1x=212x + \dfrac{1}{x} = 2\dfrac{1}{2}x+x1=221
⇒x+1x=52⇒x2+1x=52⇒2(x2+1)=5x⇒2x2+2=5x⇒2x2−5x+2=0⇒2x2−4x−x+2=0⇒2x(x−2)−1(x−2)=0⇒(x−2)(2x−1)=0⇒x−2=0 or 2x−1=0⇒x=2 or 2x=1⇒x=2 or x=12.\Rightarrow x + \dfrac{1}{x} = \dfrac{5}{2} \\[1em] \Rightarrow \dfrac{x^2 + 1}{x} = \dfrac{5}{2} \\[1em] \Rightarrow 2(x^2 + 1) = 5x \\[1em] \Rightarrow 2x^2 + 2 = 5x \\[1em] \Rightarrow 2x^2 - 5x + 2 = 0 \\[1em] \Rightarrow 2x^2 - 4x - x + 2 = 0 \\[1em] \Rightarrow 2x(x - 2) - 1(x - 2) = 0 \\[1em] \Rightarrow (x - 2)(2x - 1) = 0 \\[1em] \Rightarrow x - 2 = 0 \text{ or } 2x - 1 = 0 \\[1em] \Rightarrow x = 2 \text{ or } 2x = 1 \\[1em] \Rightarrow x = 2 \text{ or } x = \dfrac{1}{2}.⇒x+x1=25⇒xx2+1=25⇒2(x2+1)=5x⇒2x2+2=5x⇒2x2−5x+2=0⇒2x2−4x−x+2=0⇒2x(x−2)−1(x−2)=0⇒(x−2)(2x−1)=0⇒x−2=0 or 2x−1=0⇒x=2 or 2x=1⇒x=2 or x=21.
Hence, roots of the given equations are 2, 12\dfrac{1}{2}21.
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23x2−13x−1=0\dfrac{2}{3}x^2 - \dfrac{1}{3}x - 1 = 032x2−31x−1=0
6x + 29 = 5x\dfrac{5}{x}x5
3x−8x=23x - \dfrac{8}{x} = 23x−x8=2
x3+9x=4\dfrac{x}{3} + \dfrac{9}{x} = 43x+x9=4
