Question Number 10222 by Tawakalitu ayo mi last updated on 30/Jan/17

solve simultaneously. m^4 + n^4 = 9m^2 n^2 + 1 ...... (i) m + n = 4 ........ (ii)

Commented byprakash jain last updated on 01/Feb/17

Hi the font that u use appears too big. Are u using a large font size?

Answered by prakash jain last updated on 30/Jan/17

m^4 +n^4 =9m^2 n^2 +1 m^4 +n^4 +2m^2 n^2 =11m^2 n^2 +1 (m^2 +n^2 )^2 =11m^2 n^2 +1 (m^2 +n^2 +2mn−2mn)^2 =11m^2 n^2 +1 [(m+n)^2 −2mn]=11m^2 n^2 +1 (16−2mn)^2 =11m^2 n^2 +1 mn=u (16−2u)^2 =11u^2 +1 From above 1. Solve for quadratic u to get 2 values for mn. 2. Given m+n=4 and a value for mn you can find m−n and hence solve for m and n

Commented byTawakalitu ayo mi last updated on 30/Jan/17

God bless you sir.

Answered by arge last updated on 03/Feb/17

m^4 −9m^2 n^2 +n^4 −1=0 m^2 (m^2 −9n^2 )+(n^2 −1)(n^2 +1)=0 m^2 (m−3n)(m+3n)+(n^2 −1)(n^2 +1)=0......(i) m=4−n......(ii) (4−n)^2 (4−n−3n)(4−n+3n)+(n^2 −1)(n^2 +1)=0 (16−8n+n^2 )(4−4n)(4+n)+(n^2 −1)(n^2 +1)=0 4(16−8n+n^2 )(1−n)(4+n)+(n^2 −1)(n^2 +1)=0 (64−32n+4n^2 )(4+n−4n−n^2 )+(n^2 −1)(n^2 +1)=0 (64−32n+4n^2 )(4−3n−n^2 )+(n^2 −1)(n^2 +1)=0 256−192n−64n^2^ −128n+96n^2 +32n^3 +16n^2 −12n^3 −4n^4 +n^4 −1=0 3n^4 −20n^3 −48n^2 +320n−255=0 por division sintetica: (n−1)(3n^3 −17n^2 −65n+255)=0 n=1 ∴∴∴Rta m=3 ∴∴∴Rta