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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)

$$\mathrm{solve}\:\mathrm{simultaneously}. \\ $$$$\mathrm{m}^{\mathrm{4}} \:+\:\mathrm{n}^{\mathrm{4}} \:=\:\mathrm{9m}^{\mathrm{2}} \mathrm{n}^{\mathrm{2}} \:+\:\mathrm{1}\:\:\:\:\:\:\:......\:\left(\mathrm{i}\right) \\ $$$$\mathrm{m}\:+\:\mathrm{n}\:=\:\mathrm{4}\:........\:\left(\mathrm{ii}\right) \\ $$

Commented by prakash jain last updated on 01/Feb/17

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

$$\mathrm{Hi}\:\mathrm{the}\:\mathrm{font}\:\mathrm{that}\:\mathrm{u}\:\mathrm{use}\:\mathrm{appears}\:\mathrm{too} \\ $$$$\mathrm{big}.\:\mathrm{Are}\:\mathrm{u}\:\mathrm{using}\:\mathrm{a}\:\mathrm{large}\:\mathrm{font}\:\mathrm{size}? \\ $$

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

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

Commented by Tawakalitu ayo mi last updated on 30/Jan/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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

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

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