Question

(4x + 1)в квадраті - (x-5) (x+7) < 45

Answer 1

(4x+1)²-(x-5)(x+7)<45

16x²+8x+1-x²-2x+35-45<0

15x²+6x-9<0

5x²+2x-3<0

D= 4+60=64

x1= (-2+8)/10= 0,6

x2= (-2-8)/10 = -1

(x-0,6)(x+1)<0

x∈(-1;0,6)