Question Number 147487 by mnjuly1970 last updated on 21/Jul/21
![(a , 2a +1 ]∩[ a^( 2) −a , a^( 2) + 4a +1 )≠ ∅ a ∈ ?](Q147487.png)
$$ \\ $$$$ \\ $$$$\left({a}\:,\:\mathrm{2}{a}\:+\mathrm{1}\:\right]\cap\left[\:{a}^{\:\mathrm{2}} \:−{a}\:,\:{a}^{\:\mathrm{2}} +\:\mathrm{4}{a}\:+\mathrm{1}\:\right)\neq\:\varnothing \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}\:\in\:? \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 21/Jul/21
![• a<2a+1 ∧ a^2 −a<a^2 +4a+1 a>−1 ∧ a>−1/5⇒a>−1/5=−0.2 • 2a+1≥a^2 −a [∵ Intersection is empty] a^2 −3a−1≥0 a≥((3±(√(9+4)))/2)=((3±(√(13)))/2) ∵ ((3−(√(13)))/2)≯−1/5 ∴ Rejected ∴ a≥((3+(√(13)))/2) a∈[((3+(√(13)))/2),∞)](Q147551.png)
$$\bullet\:{a}<\mathrm{2}{a}+\mathrm{1}\:\wedge\:\:{a}^{\mathrm{2}} −{a}<{a}^{\mathrm{2}} +\mathrm{4}{a}+\mathrm{1}\: \\ $$$$\:\:\:\:{a}>−\mathrm{1}\:\wedge\:\:{a}>−\mathrm{1}/\mathrm{5}\Rightarrow{a}>−\mathrm{1}/\mathrm{5}=−\mathrm{0}.\mathrm{2} \\ $$$$\bullet\:\mathrm{2}{a}+\mathrm{1}\geqslant{a}^{\mathrm{2}} −{a}\:\:\left[\because\:{Intersection}\:{is}\:{empty}\right] \\ $$$$\:\:\:\:{a}^{\mathrm{2}} −\mathrm{3}{a}−\mathrm{1}\geqslant\mathrm{0} \\ $$$$\:\:\:\:{a}\geqslant\frac{\mathrm{3}\pm\sqrt{\mathrm{9}+\mathrm{4}}}{\mathrm{2}}=\frac{\mathrm{3}\pm\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\because\:\frac{\mathrm{3}−\sqrt{\mathrm{13}}}{\mathrm{2}}\ngtr−\mathrm{1}/\mathrm{5}\:\:\:\therefore\:{Rejected} \\ $$$$\:\:\:\:\:\:\:\:\therefore\:\:{a}\geqslant\frac{\mathrm{3}+\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$${a}\in\left[\frac{\mathrm{3}+\sqrt{\mathrm{13}}}{\mathrm{2}},\infty\right) \\ $$