(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


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Question Number 190731 by 073 last updated on 10/Apr/23

$({4 a}^{2} - 19 a - 5) x^{2} + a^{2} x + a + 3 = 0$ $x_{1}, x_{2} are roots$ $when, x_{1} < 0, x_{2} > 0, ∣ x_{1} ∣ - x_{2} > 0$ $interval of \max (a) = ?$ $solution ? ?$
	Answered by 073 last updated on 10/Apr/23

	Answered by manxsol last updated on 10/Apr/23
	$x_1 <0, x_2 >0 ∣x_1 ∣−x_2 >0 ⇒−x_1 −x_2 >0 x_1 +x_2 <0 −(b/a)<0 −(a^2 /((4a^2 −19a−5)))<0 (a^2 /((4a+1)(a−5)))<0 a≠0 ptos criticos + (−(1/4)) − (5) + a ε <−(1/4),5>−{0} A x_1 <0 x_2 >0 ⇒x_1 x_2 <0 (((a+3))/((4a+1)(a−5)))<0 − (−3)+ (−(1/4)) − (5) + <−∞,−3>∪<−(1/4),5> B A∩B=<−(1/4),5>−{0}$
	$x_{1} < 0, x_{2} > 0 ∣ x_{1} ∣ - x_{2} > 0$ $\Rightarrow - x_{1} - x_{2} > 0$ $x_{1} + x_{2} < 0$ $- \frac{b}{a} < 0$ $- \frac{a^{2}}{(4 a^{2} - 19 a - 5)} < 0$ $\frac{a^{2}}{(4 a + 1) (a - 5)} < 0$ $a \neq 0$ $p t o s c r i t i c o s$ $+ (- \frac{1}{4}) - (5) +$ $a ϵ < - \frac{1}{4}, 5 > - {0} A$ $x_{1} < 0 x_{2} > 0 \Rightarrow x_{1} x_{2} < 0$ $\frac{(a + 3)}{(4 a + 1) (a - 5)} < 0$ $- (- 3) + (- \frac{1}{4}) - (5) +$ $< - \infty, - 3 > \cup < - \frac{1}{4}, 5 > B$ $A \cap B =< - \frac{1}{4}, 5 > - {0}$
	Commented by073 last updated on 10/Apr/23

	$nice solution$
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