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Question Number 184438 by Shrinava last updated on 06/Jan/23

If   a − (4/( (√a))) = 17  Find   a − 4 (√a) = ?

Ifa4a=17Finda4a=?

Answered by Frix last updated on 06/Jan/23

a−(4/( (√a)))=17  Let t=(√a) ⇒ t>0  t^3 −17t−4=0  (t+4)(t−2−(√5))(t−2+(√5))=0  t>0 ⇒ t=2+(√5)  ⇒  a−4(√a)=1

a4a=17Lett=at>0t317t4=0(t+4)(t25)(t2+5)=0t>0t=2+5a4a=1

Answered by Frix last updated on 06/Jan/23

a−(4/( (√a)))=17 ⇒ a=(4/( (√a)))+17 (1)  a−4(√a)=x ⇒ a=4(√a)+x (2)  (2)−(1)  ((4a+(x−17)(√a)−4)/( (√a)))=0  4a+(x−17)(√a)−4=0  [a=4(√a)+x]  (t+4)(x−1)=0 ⇒ x=1

a4a=17a=4a+17(1)a4a=xa=4a+x(2)(2)(1)4a+(x17)a4a=04a+(x17)a4=0[a=4a+x](t+4)(x1)=0x=1

Answered by a.lgnaoui last updated on 06/Jan/23

((a(√a) −4)/( (√a)))=17  a−((4(√a))/a)=17⇒   4(√a)=a^2 −17a  a−4(√a) =18a−a^2   calcul de a  16a=(a^2 −17a)^2   (a^2 −17a−4(√a) )(a^2 −17a+4(√a) )=0  (((√a) )^3  −17((√a) )−4)((√a) )^3  −17(√a) +4)=0  (√a) =z   { ((z^3 −17z−4=0)),((z^3 −17z+4=0)) :}  .......z?   avec  a>0  ⇒   a−4(√a) =z^2 −4z

aa4a=17a4aa=174a=a217aa4a=18aa2calculdea16a=(a217a)2(a217a4a)(a217a+4a)=0((a)317(a)4)(a)317a+4)=0a=z{z317z4=0z317z+4=0.......z?aveca>0a4a=z24z

Commented by a.lgnaoui last updated on 06/Jan/23

z^3 −17z−4=0 (z=4,236067)         sit  a−4(√a) =1  z^3 −17z+4      soit  z=4        a−4(√a) =0

z317z4=0(z=4,236067)sita4a=1z317z+4soitz=4a4a=0

Commented by Frix last updated on 06/Jan/23

z=(√a) ⇔ a=z^2   z=4 ⇒ a=16  a−(4/( (√a)))=16−1=15≠17

z=aa=z2z=4a=16a4a=161=1517

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