Question

Помогите решить уравнение:
$(8x^{2} - 3x + 1)^{2} = 32x^{2} - 12x + 1$

Answer 1

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Answer 2

$displaystyle (8x^2 - 3x+1)^2 = 32x^2 - 12x+1\ \ (8x^2 - 3x+1)^2 -32x^2 +12x - 1 = 0\ \ z =8x^2 -3x \ \ -(4z+1) +(z^2 +2z+1)=0\ \ -4z-1+z^2 +2z+1=0\ \ -2z+z^2 =0\ \ -z *(2-z)=0\ \ -z = 0\ \ 2-z=0\ \ z = 0 \ z=2$

$displaystyle 8x -3 =0\ \ 8x^2-3x -2 =0 \ \ 1) x= 3/8\ \\ 2) 8x^2-3x -2 \ D= (-3)^2 - 4 *8*(-2) =73 \ \ x _1_,_2 =frac{3бsqrt{73}}{16}$

ответ: x 1 = ( 3 - sqrt(73)) / 16 , x2 =0 , x3=3/8 , x4= (3+sqrt(73))/16