Respuesta :

Answer:   " x = 0.4191113191843072889 ;  0.4191113191843072889 " .

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Step-by-step explanation:

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Given:

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 " (4x² − 1)(x+3) − (2x² − 1)(2x+1) = x² " ;  Solve for "x" ;

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First, factor:

"(4x² − 1)" ;

= 4 x² − 2x − 2x − 1 ;

→ 2 x(2x +1) − 1(2x +1) ;

= (2x − 1)(2x + 1) .

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Now, rewrite:

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" (4x² − 1)(x+3) − (2x² − 1)(2x+1) = x² " ;

by substituting for:  "(4x² − 1)" ; as follows:

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" (2x − 1)(2x + 1)(x+3) − (2x² − 1)(2x + 1) " ;

  =  " [(2x − 1)(x + 3) − (2x² − 1)] * (2x +1) = x² ;

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 Start with:

 "(2x − 1)(x + 3)" ;

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Note:  "(a + b)(c+d) = ac + ad + bc + bd " ;

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 →  (2x*x) + (2x*3)  + (-1*x) + (-1*3) ;

        =  2x² + 6x + (-1x) + (-3) ;

        =  2x² + 6x − 1x − 3 ;  

{Note: "Adding a negative is the same as "subtracting a positive."}.

            → Combine the "like terms":

                 + 6x − 1x = + 5x ;

and rewrite:

        =  2x² + 5x − 3 ;  

Now, rewrite the entire equation:

         →  [2x² + 5x − 3 − (2x² − 1)] * (2x +1) = x² ;

     

Now, consider this equation:

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         →  [2x² + 5x − 3 − 1(2x² − 1)] * (2x +1) = x² ;

   Now,  let us consider the:

          "  − 1(2x² − 1)] " ;

Take note of the "distributive property of multiplication" :

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       →  a(b + c) = ab + ac ;

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       → -1 [2x² + (-1) ] = (-1*2x²) + (-1* -1) ;

                                =  - 2x² + 1 ;

Now, we can rewrite the entire equation:

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       →  " (2x² + 5x − 3 − 2x² + 1) * (2x +1) = x² " ;

Now, consider the following part:

       →  " (2x² + 5x − 3 − 2x² + 1) " ;

          →  Combine the "like terms" :

               + 2x² − 2x² = 0 ;

               − 3 + 1 =  - 2 ;

→ and rewrite this part:

→  " ( 5x − 2) " ;

Now, we can rewrite the entire equation:

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       →  " (5x − 2) (2x + 1) = x² " ;

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Now, for the "left-hand side" of the equation:

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Note:  "(a + b)(c+d) = ac + ad + bc + bd " ;

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 →  " (5x − 2) (2x + 1) = (5x*2x) + (5x*1) +(-2*2x) + (-2*1) ;

                                 =    10x²  +  5x + (-4x)  +  (-2) ;

                                 =    10x²  +  5x  − 4x   − 2 ;

                        →  "Combine the "like terms" :

                                   + 5x  − 4x = + 1x '

                                 → and rewrite the expression:

                                 =    10x²  + 1x  − 2 ;

Now, we can rewrite the entire equation:

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             →   10x²  + 1x  − 2  =  x² ;

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Subtract:  "x² " ;  from Each Side of the equation:

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              →   10x²  + 1x  − 2 − x²  =  x² − x² ;

     to get:

              →    9x²  + 1x − 2 = 0 ; Solve for "x" ;

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      This equation is written in "quadratic format":

              →  " ax² + bx + c = 0 " ; {a ≠ 0} ;

          →  in which:  " a = 9  ; b = 1 ; c = -2 " ;

To solve for "x" using the "quadratic equation formula"

          →  x = {-b ± √(b² − 4ac) } /  {2a} ;   Plug in our known values:

              x = { -1 ± √(1² − 4(9)(-2) } /  {2*9} ;

              x = { -1 ± √[1 − (-72)} /  {18} ;

              x = { -1 ± √[1 +72} /  {18} ;

              x =  (-1  ± √73) / 18 ;

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     " x =  (-1 +√73)/18 ;  (-1 −√73) / 18 "  .

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Using calculator:

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 "  x = 0.4191113191843072889 ;  0.4191113191843072889 " .

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Hope this helps!

Good luck to you!

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