Quadratico I (Is it Quadratic?)

Algebra Level 3

40 ( 5 x + 3 ) 2 − 11 ( 5 x + 3 ) ( 3 x − 2 ) − 63 ( 3 x − 2 ) 2 4 ( x + 7 ) + 1 = 8 ( 7 x + 10 ) + 3 \frac {40(5x+3)^2-11(5x+3)(3x-2)-63(3x-2)^2}{4(x+7)+1} = 8(7x+10) + 3

Solve for x x .


The answer is 7.

Honorato Perez Iii by Honorato Perez III , 22, Philippines

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1 solution

Chew-Seong Cheong
Jun 13, 2018

40 ( 5 x + 3 ) 2 − 11 ( 5 x + 3 ) ( 3 x − 2 ) − 63 ( 3 x − 2 ) 2 4 ( x + 7 ) + 1 = 8 ( 7 x + 10 ) + 3 Let a = 5 x + 3 and b = 3 x − 2 40 a 2 − 11 a b − 63 b 2 4 x + 29 = 56 x + 83 ( 5 a − 7 b ) ( 8 a + 9 b ) 4 x + 29 = 56 x + 83 Putting back a = 5 x + 3 and b = 3 x − 2 ( 5 ( 5 x + 3 ) − 7 ( 3 x − 2 ) ) ( 8 ( 5 x + 3 ) + 9 ( 3 x − 2 ) ) 4 x + 29 = 56 x + 83 ( 4 x + 29 ) ( 67 x + 6 ) 4 x + 29 = 56 x + 83 67 x + 6 = 56 x + 83 11 x = 77 ⟹ x = 7 \begin{aligned} \frac {40(5x+3)^2-11(5x+3)(3x-2)-63(3x-2)^2}{4(x+7)+1} & = 8(7x+10) + 3 & \small \color{#3D99F6} \text{Let }a = 5x+3 \text{ and }b = 3x-2 \\ \frac {40a^2-11ab-63b^2}{4x+29} & = 56x+83 \\ \frac {(5a-7b)(8a+9b)}{4x+29} & = 56x+83 & \small \color{#3D99F6} \text{Putting back }a = 5x+3 \text{ and }b = 3x-2 \\ \frac {(5(5x+3)-7(3x-2))(8(5x+3)+9(3x-2))}{4x+29} & = 56x+83 \\ \frac {\cancel{(4x+29)}(67x+6)}{\cancel{4x+29}} & = 56x+83 \\ 67x+6 & = 56x+83 \\ 11x & = 77 \\ \implies x & = \boxed{7} \end{aligned}

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