Solve the following equation by factorisation:
x−3x+3+x+3x−3=212\dfrac{x - 3}{x + 3} + \dfrac{x + 3}{x - 3} = 2\dfrac{1}{2}x+3x−3+x−3x+3=221
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Given,
⇒x−3x+3+x+3x−3=212⇒(x−3)(x−3)+(x+3)(x+3)(x+3)(x−3)=52⇒x2−3x−3x+9+x2+3x+3x+9x2+3x−3x−9=52 ⇒x2−6x+9+x2+6x+9x2−9=52⇒2x2+18x2−9=52⇒2(2x2+18)=5(x2−9)⇒4x2+36=5x2−45⇒5x2−4x2=36+45⇒x2=81⇒x=81=±9.\phantom {\Rightarrow} \dfrac{x - 3}{x + 3} + \dfrac{x + 3}{x - 3} = 2\dfrac{1}{2} \\[1em] \Rightarrow \dfrac{(x - 3)(x - 3) + (x + 3)(x + 3)}{(x + 3)(x - 3)} = \dfrac{5}{2} \\[1em] \Rightarrow \dfrac{x^2 - 3x - 3x + 9 + x^2 + 3x + 3x + 9}{x^2 + 3x - 3x - 9} = \dfrac{5}{2} \\[1em]\ \Rightarrow \dfrac{x^2 - 6x + 9 + x^2 + 6x + 9}{x^2 - 9} = \dfrac{5}{2} \\[1em] \Rightarrow \dfrac{2x^2 + 18}{x^2 - 9} = \dfrac{5}{2} \\[1em] \Rightarrow 2(2x^2 + 18) = 5(x^2 - 9) \\[1em] \Rightarrow 4x^2 + 36 = 5x^2 - 45 \\[1em] \Rightarrow 5x^2 - 4x^2 = 36 + 45 \\[1em] \Rightarrow x^2 = 81 \\[1em] \Rightarrow x = \sqrt{81} = \pm 9.⇒x+3x−3+x−3x+3=221⇒(x+3)(x−3)(x−3)(x−3)+(x+3)(x+3)=25⇒x2+3x−3x−9x2−3x−3x+9+x2+3x+3x+9=25 ⇒x2−9x2−6x+9+x2+6x+9=25⇒x2−92x2+18=25⇒2(2x2+18)=5(x2−9)⇒4x2+36=5x2−45⇒5x2−4x2=36+45⇒x2=81⇒x=81=±9.
Hence, value of x = +9 or -9.
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4(2x - 3)2 - (2x - 3) - 14 = 0
2x2 - 9x + 10 = 0, when :
(i) x ∈ N
(ii) x ∈ Q.
4x+2−1x+3=42x+1\dfrac{4}{x + 2} - \dfrac{1}{x + 3} = \dfrac{4}{2x + 1}x+24−x+31=2x+14
5x−2−3x+6=4x\dfrac{5}{x - 2} - \dfrac{3}{x + 6} = \dfrac{4}{x}x−25−x+63=x4
