# math problem

if a2+(1/a2)=5 and a3+(1/a3)=8 then a5+ (1/a5) =?

1. Multiply both equations to get a new equation:(a2+(1/a2))•(a3+(1/a3))=5•82. To open brackets, multiply each of the terms in the first brackets with each term of the second bracketa5+(a2/a3)+(a3/a2)+(1/a5)=403. Isolate a5+(1/a5) in one side of the expressiona5+(1/a5)=40-((a3/a2)+(a2/a3))4. Simplify fractions on the right expressiona5+(1/a5)=40-(a+(1/a))
Raquel C.
16 June 2017
thanks a lot
nayeemtufat
17 June 2017
Lets solve this problem:Now assuming you'll know the expressions:a2+b2 =(a+b)2- 2ab   -------------------------- (eq. 1) a3+b3= (a+b)3-3ab(a+b)---------------------- (eq. 2) Follow the steps given by Raquel and you arrive at a5 +(1/a5) = 40 -(a+(1/a)) ,  ------------------------------- (eq. 3)To find a+1/a;a2 + 1/a2 = 5a2 + 1/a2= (a+ 1/a)2 - 2(a *1/a)               = (a + 1/a)2  - 2Which implies;(a+1/a)2  -2 = 5Isolating a + 1/a, we get (a + 1/a)2 = 7 ----------------------------------- (eq. 4)Next, a3 + 1/a3 = 8from the eqn 2;a3 + 1/a3 = (a +1/a)3 - 3( a*1/a)(a + 1/a)                = (a+1/a)3 - 3(a + 1/a)On the RHS, we will take out (a +1/a) and then interchanging LHS and RHS we get ;(a +1/a) ((a+1/a)2 -3) = a3 +1/a3(a +1/a) ((a+1/a)2 -3) = 8   ------------------- ( a3 +1/a3 = 8)(a+1/a)(7 - 3) = 8               -------------------- ( from eq.4 we know (a+ 1/a)2 = 7)(a + 1/a) (4) = 8(a + 1/a) = 8/4(a+1/a) = 2  ---------------------------------- (eq .5)put a + 1/a = 2 in eq.3We geta5 + 1/a5 = 40 -2a5 + 1/a5 = 38
Sherin T.
19 June 2017
I thought afterwards about simplifying further. I see you've already done it, Sherin. Well spoted.
Raquel C.
21 June 2017
Hello nayeemtufat,Now assuming you'll know the expressions:a2+b2 =(a+b)2- 2ab   -------------------------- (eq. 1)a3+b3= (a+b)3-3ab(a+b)---------------------- (eq. 2) Follow the steps given by Raquel and you arrive at a5 +(1/a5) = 40 -(a+(1/a)) ,  ------------------------------- (eq. 3)To find a+1/a;a2 + 1/a2 = 5a2 + 1/a2= (a+ 1/a)2 - 2(a *1/a)               = (a + 1/a)2  - 2Which implies;(a+1/a)2  -2 = 5Isolating a + 1/a, we get (a + 1/a)2 = 7 ----------------------------------- (eq. 4)Next, a3 + 1/a3 = 8from the eqn 2;a3 + 1/a3 = (a +1/a)3 - 3( a*1/a)(a + 1/a)                = (a+1/a)3 - 3(a + 1/a)On the RHS, we will take out (a +1/a) and then interchanging LHS and RHS we get ;(a +1/a) ((a+1/a)2 -3) = a3 +1/a3(a +1/a) ((a+1/a)2 -3) = 8   ------------------- ( a3 +1/a3 = 8)(a+1/a)(7 - 3) = 8               -------------------- ( from eq.4 we know (a+ 1/a)2 = 7)(a + 1/a) (4) = 8(a + 1/a) = 8/4(a+1/a) = 2  ---------------------------------- (eq .5)put a + 1/a = 2 in eq.3We geta5 + 1/a5 = 40 -2a5 + 1/a5 = 38
euticus19
06 July 2017
Hi - I know this is an old question but I thought this was a point worth making. There's actually no way to correctly evaluate a^5+1/a^5 as the two given equations are inconsistent, as you can see if you solve them individually for a (you can do this just by multiplying up and regarding them as 'disguised quadratics') - you get very different solutions. Another way to see this is by the fact that the working given below finding a+1/a=2 is correct, but if you square this equation then you get (a+1/a)^2= a^2+1/a^2+2 = 5 + 2 = 7 ~= 4 = 2^2.
Tom H.
29 July 2017
a5+1/a5= (a2+1/a2)×(a3+1/a3) -(a+1/a)              =            5   ×        8        -    root 7              = 40 - root 7;;;;                  (a+1/a)^2 - 2 =5, a+1/a= root 7
saikat1458007
28 November 2018