Question

2≤x²+x<6 розвязати ............

Answer 1

$2\leq x^2+x<6\quad \Rightarrow \; \; \left \{ {{x^2+x\geq 2} \atop {x^2+x<6}} \right.\; \left \{ {{x^2+x-2\geq 0} \atop {x^2+x-6<0}} \right. \; \left \{ {{(x-1)(x+2)\geq 0} \atop {(x-2)(x+3)<0}} \right. \\\\\left \{ {{x\in (-\infty ,-2\, ]\cup [\, 1,+\infty )} \atop {x\in (-3,2)}} \right.\; \; \Rightarrow \; \; \; x\in (-3,-2\, ]\cup [\, 1,2)\\\\\\+++++++[\, -2\, ]----[\, 1\, ]++++++\\+++(-3)-----------(2)+++$

Answer 2

x^2+x≥2

x^2+x-2≥0

D=1+8=9

x1=(-1+3)/2*1=2/2=1

x2=(-1-3)/2*1=-4/2=-2

(-oo;-2]U[1;+oo)

x^2+x<6

x^2+x-6<0

D=1+24=25

x1=(-1+5)/2*1=4/2=2

x2=(-1-5)/2*1=-6/2=-3

(-3;2)

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(-3;-2]U[1;2)

Answer 3

x^2+x≥2

x^2+x-2≥0

D=1+8=9

x1=(-1+3)/2*1=2/2=1

x2=(-1-3)/2*1=-4/2=-2

(-oo;-2]U[1;+oo)

x^2+x<6

x^2+x-6<0

D=1+24=25

x1=(-1+5)/2*1=4/2=2

x2=(-1-5)/2*1=-6/2=-3

(-3;2)

///////////////////////////////////////////////////////////////////

______-3_____________-2__________1_________2_________

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

(-3;-2]U[1;2)