Answer: " x = 0.4191113191843072889 ; 0.4191113191843072889 " .
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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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