Question

\frac { 6 x + 2 } { 4 } + \frac { 2 x ^ { 2 } - 1 } { 2 x ^ { 2 } + 2 } = \frac { 10 x - 1 } { 4 x }

Answer 1

Simplify:

Step-by-step explanation:

$\ Given \ value:\\\\\frac { 6 x + 2 } { 4 } + \frac { 2 x ^ { 2 } - 1 } { 2 x ^ { 2 } + 2 } = \frac { 10 x - 1 } { 4 x } \\\\\ Solution:\\\\\frac { 6 x + 2 } { 4 } + \frac { 2 x ^ { 2 } - 1 } { 2 x ^ { 2 } + 2 } = \frac { 10 x - 1 } { 4 x }\\\\\rightarrow \frac { (3 x + 1) } { 2 } + \frac { 2 x ^ { 2 } - 1 } { 2( x ^ { 2 } + 1) } = \frac { 10 x - 1 } { 4 x }\\\\\rightarrow \frac { (3 x + 1)( x ^ { 2 } + 1)+(2 x ^ { 2 } - 1 )} { 2( x ^ { 2 } + 1) } = \frac { 10 x - 1 } { 4 x }$

$\rightarrow \frac { (3 x^3 +3x+ x ^ { 2 } + 1)+(2 x ^ { 2 } - 1 )} { 2( x ^ { 2 } + 1) } = \frac { 10 x - 1 } { 4 x }\\\\\rightarrow \frac { 3 x^3 +3x+ x ^ { 2 } + 1+2 x ^ { 2 } - 1} { ( x ^ { 2 } + 1) } = \frac { 10 x - 1 } { 2 x }\\\\\rightarrow \frac { 3 x^3 +3x^2+ 3x } { ( x ^ { 2 } + 1) } = \frac { 10 x - 1 } { 2 x }\\\\\rightarrow (3 x^3 +3x^2+ 3x)(2 x) = (10 x - 1 )( x ^ { 2 } + 1) \\\\\rightarrow 6x^4+6x^3+6x^2 = 10x^3+10x-x^2-1\\\\\rightarrow 6x^4+6x^3+6x^2 - 10x^3-10x+x^2+1=0$

$\rightarrow 6x^4-4x^3+5x^2 +1=0$

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