Question

Решите, пожалуйста.(3b+5)x^2-2(b-1)x+2=0.

Answer 1

(3b+5)x^2 - 2(b-1)x + 2 = 0;
D = (2(b-1))^2 - 4*(3b+5)*2 = (2b-2)^2 - 8*(3b+5) = (4b^2 - 8b + 4) - 24b - 40 = 4b^2 - 8b + 4 - 24b - 40 = 4b^2 - 32b - 36;
D2 = 1024 - 4*4*(-36) = 1024 + 576 = 1600;
b1 = 72/8 = 9; b2 = -8/8 = -1;
1). (3*9+5)x^2 - 2(9-1)x + 2 = 32x^2 - 16x + 2 = 0;
D = 256 - 4*2*32 = 256 - 256 = 0;
x1 = 16/64 = 0,25;
2). (3*(-1)+5)x^2 - 2((-1)-1)x + 2 = 2x^2 + 4x + 2 = 0;
D = 16 - 4*2*2 = 0;
x2 = -4/4 = -1.
Ответ: b1 = 9; b2 = -1; x1 = 0,25; x2 = -1.
Примечание: b1 должен применяться с x1, a b2 c x2.

Answer 2

При b1 = 9, x1 = 0,25; при b2 = -1, x2 = -1.