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Question Number 193182 by York12 last updated on 06/Jun/23

  (x^2 /(2^2 −1^2 ))+(y^2 /(2^2 −3^2 ))+(z^2 /(2^2 −5^2 ))+(w^2 /(2^2 −7^2 ))=1  (x^2 /(4^2 −1^2 ))+(y^2 /(4^2 −3^2 ))+(z^2 /(4^2 −5^2 ))+(w^2 /(4^2 −7^2 ))=1  (x^2 /(6^2 −1^2 ))+(y^2 /(6^2 −3^2 ))+(z^2 /(6^2 −5^2 ))+(w^2 /(6^2 −7^2 ))=1  (x^2 /(8^2 −1^2 ))+(y^2 /(8^2 −3^2 ))+(z^2 /(8^2 −5^2 ))+(w^2 /(8^2 −7^2 ))=1     find (x^2 +y^2 +z^2 +w^2 ).

x22212+y22232+z22252+w22272=1x24212+y24232+z24252+w24272=1x26212+y26232+z26252+w26272=1x28212+y28232+z28252+w28272=1find(x2+y2+z2+w2).

Answered by York12 last updated on 06/Jun/23

  Let s takes the values (4 , 16 , 36 , 64)  ∴ (x^2 /(s−1))+(y^2 /(s−9))+(z^2 /(s−25))+(w^2 /(s−49))=1  ⇒ x^2 (s−9)(s−25)(s−49) + y^2 (s−1)(s−25)(s−49)+z^2 (s−1)(s−9)(s−49)+w^2 (s−1)(s−9)(s−25)=(s−1)(s−9)(s−25)(s−49)  ⇒x^2 (s−9)(s−25)(s−49) + y^2 (s−1)(s−25)(s−49)+z^2 (s−1)(s−9)(s−49)+w^2 (s−1)(s−9)(s−25)−(s−1)(s−9)(s−25)(s−49)=0  let the previous expression be a quartic  polynomial in s with leading coefficient of (−1)  and we know the values of s for which that  holds s ∈ { 4 , 16 , 36 , 64 }    Then :   x^2 (s−9)(s−25)(s−49) + y^2 (s−1)(s−25)(s−49)+z^2 (s−1)(s−9)(s−49)+w^2 (s−1)(s−9)(s−25)−(s−1)(s−9)(s−25)(s−49)=  −(s−4)(s−16)(s−36)(s−64) = f(s)  f(1)= −(3×15×35×63)=−x^2 (8×24×48)  ⇒x^2 =((3×15×35×63)/(8×24×48)) .......(i)  f(9)=(5×7×27×55)=y^2 (8×16×40)   ⇒ y^2 =(((5×7×27×55))/((8×16×40))) .......(ii)  f(25)=−(21×9×11×39)=−z^2 (24×16×24)  ⇒ z^2 =(((21×9×11×39))/((24×16×24)))......(iii)  f(49)=(45×33×13×15)=w^2 (48×40×24)  ⇒w^2 =(((45×33×13×15))/((48×40×24))) .......(iv)  (i)+(ii)+(iii)+(iv)= 36 ★

Letstakesthevalues(4,16,36,64)x2s1+y2s9+z2s25+w2s49=1x2(s9)(s25)(s49)+y2(s1)(s25)(s49)+z2(s1)(s9)(s49)+w2(s1)(s9)(s25)=(s1)(s9)(s25)(s49)x2(s9)(s25)(s49)+y2(s1)(s25)(s49)+z2(s1)(s9)(s49)+w2(s1)(s9)(s25)(s1)(s9)(s25)(s49)=0letthepreviousexpressionbeaquarticpolynomialinswithleadingcoefficientof(1)andweknowthevaluesofsforwhichthatholdss{4,16,36,64}Then:x2(s9)(s25)(s49)+y2(s1)(s25)(s49)+z2(s1)(s9)(s49)+w2(s1)(s9)(s25)(s1)(s9)(s25)(s49)=(s4)(s16)(s36)(s64)=f(s)f(1)=(3×15×35×63)=x2(8×24×48)x2=3×15×35×638×24×48.......(i)f(9)=(5×7×27×55)=y2(8×16×40)y2=(5×7×27×55)(8×16×40).......(ii)f(25)=(21×9×11×39)=z2(24×16×24)z2=(21×9×11×39)(24×16×24)......(iii)f(49)=(45×33×13×15)=w2(48×40×24)w2=(45×33×13×15)(48×40×24).......(iv)(i)+(ii)+(iii)+(iv)=36

Commented by manxsol last updated on 06/Jun/23

dominio de f(s)={4,16,36,64}  evaluation={1,9,25,49}   how do you explanation?

dominiodef(s)={4,16,36,64}evaluation={1,9,25,49}howdoyouexplanation?

Commented by York12 last updated on 07/Jun/23

{4,16,36,64} that is the solution set   for which f(s)=0  {1,9,25,49} those are just values I have   plugged them into the function to   eliminate x^2  , y^2  , z^2  , w^2   so that I can obtain their values

{4,16,36,64}thatisthesolutionsetforwhichf(s)=0{1,9,25,49}thosearejustvaluesIhavepluggedthemintothefunctiontoeliminatex2,y2,z2,w2sothatIcanobtaintheirvalues

Commented by manxsol last updated on 07/Jun/23

thank for explanation.

thankforexplanation.

Commented by York12 last updated on 07/Jun/23

you are welcome

youarewelcome

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