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a-24-a-3-b-2-a-2a-b-2-a-b-2-20-would-you-solve-this-

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Question Number 5159 by 1771727373 last updated on 24/Apr/16
a+24((a−3)/(b^2 +a))=2a−b^(2 ) a×b^2 =20 would you solve this?
a+24a−3b2+a=2a−b2a×b2=20wouldyousolvethis?
Answered by FilupSmith last updated on 24/Apr/16
ab^2 =20 ⇒ b^2 =20a ∴ a+24((a−3)/(b^2 +a))=2a−b^(2 ) ⇒ a+24((a−3)/(((20)/a)+a))=2a−((20)/a) a+24((a−3)/((20+a^2 )/a))=((2a^2 −20)/a) a+24((a^2 −3a)/(20+a^2 ))=((2a^2 −20)/a) a^2 +24a(a^2 −3a)=(2a^2 −20)(20+a^2 ) a^2 +24a^3 −72a=40a^2 +2a^4 −20^2 −20a^2 −19a^2 +24a^3 −2a^4 −72a+400=0 19a^2 −24a^3 +2a^4 +72a−400=0 2a^4 −24a^3 +19a^2 +72a−400=0 continue
ab2=20⇒b2=20a∴a+24a−3b2+a=2a−b2⇒a+24a−320a+a=2a−20aa+24a−320+a2a=2a2−20aa+24a2−3a20+a2=2a2−20aa2+24a(a2−3a)=(2a2−20)(20+a2)a2+24a3−72a=40a2+2a4−202−20a2−19a2+24a3−2a4−72a+400=019a2−24a3+2a4+72a−400=02a4−24a3+19a2+72a−400=0continue
Answered by Yozzii last updated on 28/Apr/16
a(b^2 +a)+24(a−3)=(2a−b^2 )(b^2 +a) ab^2 +a^2 +24a−72=2ab^2 +2a^2 −b^4 −ab^2 b^4 −a^2 +24a−72=0 b^4 =a^2 −24a+72 b^4 =((400)/a^2 ) ∴ 400=a^4 −24a^3 +72a^2 a^4 −24a^3 +72a^2 −400=0 (∗) Let f(x)=x^4 −24x^3 +72x^2 −400. By trial and error, and application of the intermediate value theorem, a root for the equation exists in the interval [20.5,20.6]. f^′ (x)=4x^3 −72x^2 +144x By the Newton−Raphson method the n+1 th approximation to the root of f(x)=0, given the n th approximation, is found by x_(n+1) =x_n −((f(x_n ))/(f^′ (x_n ))) x_(n+1) =((3x_n ^4 −48x_n ^3 +72x_n ^2 +400)/(4x_n ^3 −72x_n ^2 +144x_n )) Let x_1 =((20.5+20.6)/2)=20.55. ⇒x_2 ≈20.540974 ⇒x_3 ≈20.540961 ⇒x_4 ≈20.540961 ∴ a≈20.541. Note that (∗) has more than just one real root. This same method could be used to search for other real roots. Since b^2 =((20)/a), if b∈R⇒a>0 ∴ b=±(√((20)/(20.541)))≈±0.9867
a(b2+a)+24(a−3)=(2a−b2)(b2+a)ab2+a2+24a−72=2ab2+2a2−b4−ab2b4−a2+24a−72=0b4=a2−24a+72b4=400a2∴400=a4−24a3+72a2a4−24a3+72a2−400=0(∗)Letf(x)=x4−24x3+72x2−400.Bytrialanderror,andapplicationoftheintermediatevaluetheorem,arootfortheequationexistsintheinterval[20.5,20.6].f′(x)=4x3−72x2+144xBytheNewton−Raphsonmethodthen+1thapproximationtotherootoff(x)=0,giventhenthapproximation,isfoundbyxn+1=xn−f(xn)f′(xn)xn+1=3xn4−48xn3+72xn2+4004xn3−72xn2+144xnLetx1=20.5+20.62=20.55.⇒x2≈20.540974⇒x3≈20.540961⇒x4≈20.540961∴a≈20.541.Notethat(∗)hasmorethanjustonerealroot.Thissamemethodcouldbeusedtosearchforotherrealroots.Sinceb2=20a,ifb∈R⇒a>0∴b=±2020.541≈±0.9867

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