Question

Să se rezolve inecuațiile:
a) $x^{2}$-3x+4≥3x+2;
b) 2$x^{2}$-2x<-$\frac{1}{2}$;
c) -$x^{2}$-3x+5<$x^{2}$-1;
d) $x^{2}$+x+7≤0;
e) -$x^{2}$+2x>x+$\frac{1}{4}$.

Answer 1

1) x²-3x+4≥3x+2
x²-3x+4-3x-2≥0
x²-6x+2≥0
x²-6x+2=0
x=3+√7
x=3-√7
1(x-(3+√7))·(x-(3-√7))≥0
(x-3-√7)·(x-3+√7)≥0
$\left \{ {{x-3- \sqrt{7} } \geq 0\atop {x-3+ \sqrt{7} \geq 0 }} \right.$

$\left \{ {{x-3- \sqrt{7} \leq 0 } \atop {x-3+ \sqrt{7} \leq 0 }} \right.$

$\left \{ {{x \geq 3+ \sqrt{7} } \atop {x=3- \sqrt{7} }} \right.$

x∈[3+√7,+∞}
x∈{-∞,3-√7]
x∈{-∞,3-√7]∪[3+√7,+∞}

3) x²+x+7≤0
x²+x+7=0
x∉R
x²+x+7≤0,a=1
x∈∅
succes!!!!