Résoudre une inéquation du second degré sans terme nul - 1ère - Exercice Mathématiques

2x^2+3x-1\geqslant 3x^2-3x+8

S=\left\{ 3 \right\}

S=\left[3; +\infty \right[

S=\left]-\infty;3 \right]

S=\varnothing

2x\left(x-3\right)\leqslant \left(3x+1\right)^2

S=\left] -\infty;\dfrac{-6-\sqrt{29}}{7} \right]\cup\left[ \dfrac{-6+\sqrt{29}}{7};+\infty \right[

S=\left] \dfrac{-6-\sqrt{29}}{7};\dfrac{-6+\sqrt{29}}{7}\right[

S=\left] -\infty;\dfrac{-6-\sqrt{29}}{7} \right[\cup\left] \dfrac{-6+\sqrt{29}}{7};+\infty \right[

S=\varnothing

-3x^2+8x-2\leqslant -2x^2+6x-5

S=\left] -\infty;-1 \right]\cup\left[3;+\infty \right[

S=\left] -\infty;-1 \right[\cup\left]3;+\infty \right[

S=\left[ -1 ; 3 \right]

S=\left]-1;3 \right[

4x^2-x+2\gt x^2+5x

S=\left] -\infty;\dfrac{3-\sqrt{3}}{3} \right[\cup\left] \dfrac{3+\sqrt{3}}{3};+\infty \right[

S=\left] -\infty;\dfrac{3-\sqrt{3}}{3} \right]\cup\left[ \dfrac{3+\sqrt{3}}{3};+\infty \right[

S=\left] \dfrac{3-\sqrt{3}}{3}; \dfrac{3+\sqrt{3}}{3} \right[

S=\varnothing

3x^2-x+2\lt -x^2+2x+4

S=\left] \dfrac{3-\sqrt{41}}{8};\dfrac{3+\sqrt{41}}{8} \right[

S=\left[ \dfrac{3-\sqrt{41}}{8};\dfrac{3+\sqrt{41}}{8} \right]

S=\left] -\infty; \dfrac{3-\sqrt{41}}{8} \right[ \cup\left] \dfrac{3+\sqrt{41}}{8};+\infty \right[

S=\varnothing