Question

The equation is attached.
Solve for x:

Answer 1

$\large \bf {\dfrac{x^{2} + 1}{2} - \dfrac{x^{2} - 1}{4} = \dfrac{x+1}{8}}\\\\\\Taking \: LCM,\\\\\\\leadsto \sf{\dfrac{2(x^{2} + 1) - x^{2} - 1}{4} = \dfrac{x+1}{8}}\\\\\\\\\dashrightarrow \: \sf{\dfrac{2x^{2} +2 - x^{2} - 1}{4} = \dfrac{x+1}{8}}\\\\\\\\\implies \sf{\dfrac{x^{2} + 1}{4} = \dfrac{x+1}{8}}\\\\\\ \tt{On \: Cross \: Multiplication,}\\\\\\\rightarrow \sf{4(x + 1) = 8(x^{2} +1)} \\\\\\\implies \sf{4x + 4 = 8x^{2} +8}\\\\\\\implies \sf{8x^{2} + 8 = 4x + 4}$

$\implies \sf{8x^{2} + 8 - 4 = 4x}\\\\\\\implies \sf{8x^{2} + 4 = 4x }\\\\Dividing \: by \: 4 \: on \: both \: sides\\\\\\\implies \sf{2x^{2} + 1 = x}\\\\\implies \sf{2x^{2} - x + 1 = 0}\\\\\implies \sf{2x^{2} -2x + x + 1 = 0}\\\\\implies \sf{2x(x - 1) + 1(x - 1) = 0}\\\\\implies \sf{(2x+ 1)(x-1)=0}\\\\\sf{(2x+1) \: and \: (x - 1) \: are \: the \: factors}$

$\implies \sf{(2x + 1) = 0 \quad \quad OR \quad \quad (x - 1) = 0}\\\\\\\\\implies \boxed{\large{\sf{\boxed{\sf{x = \dfrac{-1}{2} \quad or \quad x = 1}}}}}$