Question

Помогите срочно! (x+8)^2+1=0 (4x+8)^2-12=0 (3-2x)^2=36 (2+11x)^2=6

Answer 1

1)(x+8)^2+1=0
(x+8)^2=-1
x^2+16x+65=-1
x^2+16x+66=0
x ∉ R

2)(4x+8)^2-12=0
16x^2+64x+64-12=0
16x^2+64x+52=0

x₁=$\frac{-4- \sqrt{3} }{2}$ x₂=$\frac{-4+ \sqrt{3} }{2}$

3)(3-2x)^2=36
9-2x+4x^2=36
4x^2-12x-27=0

x₁= -$\frac{3}{2}$  x₂=$\frac{9}{2}$

4)(2+11x)^2=6121x^2+44x+4=6
121x^2+44x-2=0

x₁=$\frac{-2+ \sqrt{6} }{11}$  x₂=$\frac{-2- \sqrt{6} }{11}$