Use componendo and dividendo to find the value of x, when :
x2+3x3x2+1=1413\dfrac{x^2 + 3x}{3x^2 + 1} = \dfrac{14}{13}3x2+1x2+3x=1314
Given,
Applying componendo and dividendo, we get :
⇒x2+3x+3x2+1x2+3x−(3x2+1)=14+1314−13⇒x2+3x+3x2+1x2+3x−3x2−1=271⇒(x+1)3(x−1)3=3313⇒(x+1x−1)3=(31)3⇒x+1x−1=3⇒x+1=3(x−1)⇒x+1=3x−3⇒3x−x=1+3⇒2x=4⇒x=42=2.\Rightarrow \dfrac{x^2 + 3x + 3x^2 + 1}{x^2 + 3x - (3x^2 + 1)} = \dfrac{14 + 13}{14 - 13} \\[1em] \Rightarrow \dfrac{x^2 + 3x + 3x^2 + 1}{x^2 + 3x - 3x^2 - 1}= \dfrac{27}{1} \\[1em] \Rightarrow \dfrac{(x + 1)^3}{(x - 1)^3} = \dfrac{3^3}{1^3} \\[1em] \Rightarrow \Big(\dfrac{x + 1}{x - 1}\Big)^3 = \Big(\dfrac{3}{1}\Big)^3 \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = 3 \\[1em] \Rightarrow x + 1 = 3(x - 1) \\[1em] \Rightarrow x + 1 = 3x - 3 \\[1em] \Rightarrow 3x - x = 1 + 3 \\[1em] \Rightarrow 2x = 4 \\[1em] \Rightarrow x = \dfrac{4}{2} = 2.⇒x2+3x−(3x2+1)x2+3x+3x2+1=14−1314+13⇒x2+3x−3x2−1x2+3x+3x2+1=127⇒(x−1)3(x+1)3=1333⇒(x−1x+1)3=(13)3⇒x−1x+1=3⇒x+1=3(x−1)⇒x+1=3x−3⇒3x−x=1+3⇒2x=4⇒x=24=2.
Hence, x = 2.
