Math Question and Answers


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 177042 by Ar Brandon last updated on 30/Sep/22

	Answered by a.lgnaoui last updated on 30/Sep/22
	$somme dds racinesx_1 +x_2 =−a^2 x_1 x_2 =((a+3)/(4a^2 −19a−5)) Δ=a^4 −4(4a^2 −19a−5)(a+3)>0 a^4 −4(4a^3 +12a^2 −19a^2 −54a−5a−15) a^4 −4(4a^3 −7a^2 −59a−15) Δ=(a+2,714)(a+0,263)(a−6,991)(a−11,986) a^4 −16a^3 +28a^2 +236a+60) x_1 =(−a^2 ±(√(a^4 −16a^3 +28a^2 +236a+60)) ) /2 = x_1 =((−a^2 −(√(a^4 −16a^3 +28a^2 +236a+60)))/2) a=0 a∈{−(1/4),5} x_1 =−((√(60))/2)<0 x_2 >0 ∣x ∣−x_2 =(√(60)) >0 a∈]−(1/4) ,5[ satisfait •a=−1 pas de solutions reels donc a∉]−3,−(1/4)[ •a= •a=−4 a=−4∈{−∞,−3} x1=−8−((√(784)) )/2<0 ∣ x_2 =−8+((√(784))) /2>0 ∣x1∣−x2 >0 • donc a∈] −∞,−3[ satisfait aux conditions a=−3 x1=−(9/2)−(√(81+656+84−678+60 )) )/2=((−9)/2)−(√(113))<0(λ>0) x_2 =−(9/2)+(√(113))>0 ∣x1∣−x2=9>0 • donc a=−3 satisfait a=−2 x_1 =−2−(√(16+128+112−472+60)) pas de solutiins reels dinc a∉]−3,−(1/4)[ •a=5 x1<0 ∣x1−x2>0 a=5 satidfait •a=6 x1=−18−((√(2344)) /2) <0 x2=−18+((√(2344)) /2>0 ∣x1∣=18+((√(2344)) /2)+18−((√(2344)) /2)>0 • donc a∈[5,+∞ [ satisfait satisfait aux conditions •a=−(1/4) x1=−(1/(32))−(((13)/(3(√7))) )/2 x_2 =−(1/(32))+((13)/(3(√7))) ∣x1∣−x2>0 •a=−(1/4) satisfait conclusion: a∈]−∞,−3]∪ [−(1/4),5]∪[5,+∞[$
	$s o m m e d d s {r a c i n e s x}_{1} + x_{2} = - a^{2} x_{1} x_{2} = \frac{a + 3}{4 a^{2} - 19 a - 5}$ $Δ = a^{4} - 4 (4 a^{2} - 19 a - 5) (a + 3) > 0$ $a^{4} - 4 (4 a^{3} + 12 a^{2} - 19 a^{2} - 54 a - 5 a - 15)$ $a^{4} - 4 (4 a^{3} - 7 a^{2} - 59 a - 15)$ $Δ = (a + 2, 714) (a + 0, 263) (a - 6, 991) (a - 11, 986)$ $a^{4} - 16 a^{3} + 28 a^{2} + 236 a + 60)$ $x_{1} = (- a^{2} \pm \sqrt{a^{4} - 16 a^{3} + 28 a^{2} + 236 a + 60}) / 2$ $=$ $x_{1} = \frac{- a^{2} - \sqrt{a^{4} - 16 a^{3} + 28 a^{2} + 236 a + 60}}{2}$ $a = 0 a \in {- \frac{1}{4}, 5} x_{1} = - \frac{\sqrt{60}}{2} < 0 x_{2} > 0$ $∣ x ∣ - x_{2} = \sqrt{60} > 0$ $a \in] - \frac{1}{4}, 5 [s a t i s f a i t$ $∙ a = - 1 p a s d e s o l u t i o n s r e e l s$ $d o n c a \notin] - 3, - \frac{1}{4} [$ $∙ a =$ $∙ a = - 4$ $a = - 4 \in {- \infty, - 3} x 1 = - 8 - (\sqrt{784}) / 2 < 0 ∣$ $x_{2} = - 8 + (\sqrt{784)} / 2 > 0$ $∣ x 1 ∣ - x 2 > 0$ $∙ d o n c a \in] - \infty, - 3 [s a t i s f a i t a u x c o n d i t i o n s$ $a = - 3 x 1 = - \frac{9}{2} - \sqrt{81 + 656 + 84 - 678 + 60}) / 2 = \frac{- 9}{2} - \sqrt{113} < 0 (λ > 0)$ $x_{2} = - \frac{9}{2} + \sqrt{113} > 0 ∣ x 1 ∣ - x 2 = 9 > 0$ $∙ d o n c a = - 3 s a t i s f a i t$ $a = - 2 x_{1} = - 2 - \sqrt{16 + 128 + 112 - 472 + 60} p a s d e s o l u t i i n s r e e l s$ $d i n c a \notin] - 3, - \frac{1}{4} [$ $∙ a = 5 x 1 < 0 ∣ x 1 - x 2 > 0$ $a = 5 s a t i d f a i t$ $∙ a = 6 x 1 = - 18 - (\sqrt{2344} / 2) < 0$ $x 2 = - 18 + (\sqrt{2344} / 2 > 0$ $∣ x 1 ∣= 18 + (\sqrt{2344} / 2) + 18 - (\sqrt{2344} / 2) > 0$ $∙ d o n c a \in [5, + \infty [s a t i s f a i t$ $s a t i s f a i t a u x c o n d i t i o n s$ $∙ a = - \frac{1}{4} x 1 = - \frac{1}{32} - (\frac{13}{3 \sqrt{7}}) / 2$ $x_{2} = - \frac{1}{32} + \frac{13}{3 \sqrt{7}} ∣ x 1 ∣ - x 2 > 0$ $∙ a = - \frac{1}{4} s a t i s f a i t$ $c o n c l u s i o n :$ $a \in] - \infty, - 3] \cup [- \frac{1}{4}, 5] \cup [5, + \infty [$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum