Răspuns:
Explicație pas cu pas:
1) z = a+bi, z barat a -bi
A = (a+bi)(2+3i) + (a-bi)(2-3i) =
2a +3ai +2bi -3b + 2a -3i -2bi -3b =
4a - 6b este in R
2) x^2 -6x + 7 = 0, delta = 36-28 = 8
x1,2 = (6 +- 2√2)/2
x1 = 3 -√2, x2 = 3 +√2
f(3 +√2) = 0 (e radacina), deci produsul = 0
3) x^2 +x -2 >0, (x+2)(x-1) > 0
Radacini: -2 si 1
In afara rad. semn diferit de a, x^2 +x -2 <0
Asadar x^2 +x -2 >0 pt. x in afara rad. ,
adica x in (-inf, -2)U(1, +inf)
(x-1)/2 > 0, x > 1
Deci cautam radacini in (1, +inf)
lg(x^2 +x -2) = lg10 + lg((x-1)/2)
lg(x^2 +x -2 ) = lg(10*(x-1)/2) = lg(5(x-1))
x^2 +x -2 = 5x -5
x^2 -4x + 3 = 0, delta = 16 -12 = 4
x1,2 = (4 +- 2)/2
x1 = 2/2 = 1 nu convine
x2 = 6/2 = 3 convine
Cam atat...
