Question

Resolva em R ,a inequação
${x}^{2} + 3x \leqslant 2 {x }^{2} - 1$
urgente!​

Answer 1

Explicação passo-a-passo:

$x^{2} +3x\leq 2x^2-1\\ \\ x^{2} -2x^2+3x+1\leq 0\\ \\ -x^2+3x+1\leq 0$

a=-1

b=3

c=1

Δ=b²-4ac

Δ=3²-4(-1)(1)

Δ=9+4

Δ=13

$x={-b\pm\sqrt{\Delta} \over2a}={-3\pm\sqrt{13} \over2(-1)}={-3\pm\sqrt{13} \over-2}={-1\over-2}(3\pm\sqrt{13} )={3\pm\sqrt{13} \over2}\\ \\ x_1={3-\sqrt{13} \over2}\\ \\ x_2={3+\sqrt{13}\over2}\\  \\ Esquema\\ \\ ----\bullet^{{3-\sqrt{13} \over2}}++++\bullet^{{3+\sqrt{13} \over2}}----\\ ~~~~~m/a~~~~~~~~~~~~~~c/a~~~~~~~~~~~~~~~~~m/a$

Como pediu ≤

$S=\{x\in R/x\leq {3-\sqrt{13} \over2}~~ ou~~x\geq {3+\sqrt{13} \over2}\}$