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4a-2-19a-5-x-2-a-2-x-a-3-0-x-1-x-2-are-roots-when-x-1-lt-0-x-2-gt-0-x-1-x-2-gt-0-interval-of-max-a-solution-

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Question Number 190731 by 073 last updated on 10/Apr/23
(4a^2 −19a−5)x^2 +a^2 x+a+3=0 x_1 ,x_2 are roots when , x_1 <0 ,x_2 >0 , ∣x_1 ∣−x_2 >0 interval of max(a)=? solution??
$$\left(\mathrm{4a}^{\mathrm{2}} −\mathrm{19a}−\mathrm{5}\right)\mathrm{x}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} \mathrm{x}+\mathrm{a}+\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{x}_{\mathrm{1}} ,\mathrm{x}_{\mathrm{2}} \mathrm{are}\:\mathrm{roots} \\ $$$$\mathrm{when}\:,\:\mathrm{x}_{\mathrm{1}} <\mathrm{0}\:\:\:,\mathrm{x}_{\mathrm{2}} >\mathrm{0}\:\:,\:\mid\mathrm{x}_{\mathrm{1}} \mid−\mathrm{x}_{\mathrm{2}} >\mathrm{0} \\ $$$$\mathrm{interval}\:\mathrm{of}\:\:\:\mathrm{max}\left(\mathrm{a}\right)=? \\ $$$$\mathrm{solution}?? \\ $$
Answered by 073 last updated on 10/Apr/23
Answered by manxsol last updated on 10/Apr/23
x_1 <0, x_2 >0 ∣x_1 ∣−x_2 >0 ⇒−x_1 −x_2 >0 x_1 +x_2 <0 −(b/a)<0 −(a^2 /((4a^2 −19a−5)))<0 (a^2 /((4a+1)(a−5)))<0 a≠0 ptos criticos + (−(1/4)) − (5) + a ε <−(1/4),5>−{0} A x_1 <0 x_2 >0 ⇒x_1 x_2 <0 (((a+3))/((4a+1)(a−5)))<0 − (−3)+ (−(1/4)) − (5) + <−∞,−3>∪<−(1/4),5> B A∩B=<−(1/4),5>−{0}
$${x}_{\mathrm{1}} <\mathrm{0},\:{x}_{\mathrm{2}} >\mathrm{0}\:\:\:\:\:\:\mid{x}_{\mathrm{1}} \mid−{x}_{\mathrm{2}} >\mathrm{0} \\ $$$$\Rightarrow−{x}_{\mathrm{1}} −{x}_{\mathrm{2}} >\mathrm{0} \\ $$$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} <\mathrm{0} \\ $$$$−\frac{{b}}{{a}}<\mathrm{0} \\ $$$$−\frac{{a}^{\mathrm{2}} }{\left(\mathrm{4}{a}^{\mathrm{2}} −\mathrm{19}{a}−\mathrm{5}\right)}<\mathrm{0} \\ $$$$\frac{{a}^{\mathrm{2}} }{\left(\mathrm{4}{a}+\mathrm{1}\right)\left({a}−\mathrm{5}\right)}<\mathrm{0} \\ $$$${a}\neq\mathrm{0} \\ $$$${ptos}\:{criticos} \\ $$$$\:\:\:\:\:\:\:+\:\:\:\:\:\:\left(−\frac{\mathrm{1}}{\mathrm{4}}\right)\:\:\:\:\:\:−\:\:\:\:\:\left(\mathrm{5}\right)\:\:\:\:\:\:+ \\ $$$${a}\:\epsilon\:\:\:<−\frac{\mathrm{1}}{\mathrm{4}},\mathrm{5}>−\left\{\mathrm{0}\right\}\:\:\:\:\:\:{A} \\ $$$${x}_{\mathrm{1}} <\mathrm{0}\:{x}_{\mathrm{2}} >\mathrm{0}\:\:\:\:\Rightarrow{x}_{\mathrm{1}} {x}_{\mathrm{2}} <\mathrm{0} \\ $$$$\frac{\left({a}+\mathrm{3}\right)}{\left(\mathrm{4}{a}+\mathrm{1}\right)\left({a}−\mathrm{5}\right)}<\mathrm{0} \\ $$$$−\:\left(−\mathrm{3}\right)+\:\:\:\:\:\left(−\frac{\mathrm{1}}{\mathrm{4}}\right)\:\:\:\:\:−\:\:\:\:\left(\mathrm{5}\right)\:\:\:+ \\ $$$$<−\infty,−\mathrm{3}>\cup<−\frac{\mathrm{1}}{\mathrm{4}},\mathrm{5}>\:\:{B} \\ $$$${A}\cap{B}=<−\frac{\mathrm{1}}{\mathrm{4}},\mathrm{5}>−\left\{\mathrm{0}\right\}\:\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by 073 last updated on 10/Apr/23
nice solution
$$\mathrm{nice}\:\mathrm{solution} \\ $$

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