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avatar+289 

What is the greatest possible value of x for the equation

 

\((4x-16/3x-4)^2 + (4x-16/3x-4) = 12\)

 Jan 9, 2024

Best Answer 

 #1
avatar+36925 
+1

Expand the far left side, then multiply the second term by   (3x-4)/3x-4)   and multiply the entire equation by  ( 3x-4)^2 

  and simplify the mess to get 

          80x^2- 96-128 = 0        Quadrtaic formula then shows x = 2 as the largest x 

 Jan 9, 2024
 #1
avatar+36925 
+1
Best Answer

Expand the far left side, then multiply the second term by   (3x-4)/3x-4)   and multiply the entire equation by  ( 3x-4)^2 

  and simplify the mess to get 

          80x^2- 96-128 = 0        Quadrtaic formula then shows x = 2 as the largest x 

ElectricPavlov Jan 9, 2024
 #2
avatar+397 
+1

Much easier if you notice that this is a quadratic X^2 + X - 12= 0, where X = (4x -16)/(3x - 4).

Solve for X (= -4 or 3 ) and then for x.

 Jan 9, 2024

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