Question

A)fie A multimea solutiilor reale a ecuatiei 2x²+5x-3=0
determinati multimea A∩{-3;3}
b)fie A multimea solutiilor reale a ecuatiei 6x²-5x-1=0
determinati multimea A∩{-1;1}

Answer 1

$2 {x}^{2} + 5x - 3 = 0 \\ x = \frac{ - 5 + - \sqrt{ {5}^{2} - 4 \times 2 \times - 3} }{2 \times 2} \\ x = \frac{ - 5 + - \sqrt{49} }{4} \\ x = \frac{ - 5 + - 7}{4} \\ = > x1 = \frac{ - 5 + 7}{4} = > x1 = \frac{1}{2} \\ = > x2 = \frac{ - 5 - 7}{4} = > x2 = - 3$
A = { 1/2 ; - 3 }
Si atunci A intersectat {-3 ; 3 } = {-3}

A doua se rezolva cu delta exact la fel ca cea de mai sus asa ca o rezolv altfel :)
${6x}^{2} - 5x - 1 = 0 \\ 6 {x}^{2} + x - 6x - 1 = 0 \\ x(6x + 1) - (6x + 1) = 0 \\ (6x + 1)(x - 1) = 0 \\ = > 6x + 1 = 0 = > x1 = - \frac{1}{6} \\ = > x - 1 = 0 = > x2 = 1$
Deci A = { - 1/6 ; 1 }
A intersectat cu { - 1 ; 1 } = { 1 }