Question

1)x^2-4x>-4
2)(2x-1)^2<4x+61

Answer 1

х^2-4x>-4
x^2-4x+4>0
x^2-4x+4=0
по т.Виета
x*x = 4    ⇒   x=2
x+x = 4         x=2
отв.: x∈ (2;+∞)

Answer 2

1) $x^{2} -4x > -4 \\ x^{2} -4x +4>0 \\ D=16-16=0 \\ x> \frac{4}{2} \\ x>2$

2) $(2x-1)^{2} <4x+61 \\ 4 x^{2} -4x+1-4x-61<0 \\ 4 x^{2} - 8x - 60<0 \\ x^{2} -2x-15<0 \\ D=4+60=64 \\ \sqrt{64} =8 \\ x_{1} = \frac{2+8}{2} =5 \\ x_{2} = \frac{2-8}{2} = -3 \\ x_{1}<5 \\ x_{2} <-3$