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Solve the equations x^4+y^4=257, x+y=5.

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sixpenpencil | Student, Grade 10 | eNoter

Posted August 16, 2010 at 1:51 AM via web

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Solve the equations x^4+y^4=257, x+y=5.

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giorgiana1976 | College Teacher | Valedictorian

Posted August 16, 2010 at 1:54 AM (Answer #1)

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For the beginning, we'll note:

x = a+b

y = a-b

So, x^4 + y^4 = (a+b)^4 + (a-b)^4 = 257

We'll remove the brackets:

a^4 + 4a^3*b+ 6a^2*b^2+4ab^3+b^4+a^4-4a^3*b+6a^2*b^2-4ab^3 + b^4 = 257

We'll eliminate like terms, we'll factorize and we'll get:

2(a^4+6a^2*b^2+b^4) = 257

a^4+6a^2*b^2+b^4= 257/2

We also know, from enunciation, that x+y=5

We'll substitute x and y by a+b and a-b:

a+b+a-b = 5

We'll eliminate like terms:

2a = 5

a = 5/2

We'll substitute the value of a in the relation

a^4+6a^2*b^2+b^4= 257/2

(5/2)^4 + 6*(5/2)^2*b^2 + b^4 = 257/2

b^4 + (75/2)*b^2 + 625/16 = 257/2

16b^4 + 600b^2 - 1431 = 0

b^2 = (-600+672)/32

b^2 = 9/4

b1 = +sqrt(9/4)

b1 = 3/2

b2 = -3/2

b^2 = -159/4

b3 = i*sqrt159/2

b4 = -i*sqrt159/2

The solutions are:

-for a = 5/2 and b = 3/2

x = a+b = 4

y = a-b = 1

-for a = 5/2 and b = -3/2

x = 1

y = 4

-for a = 5/2 and b = i*sqrt159/2

x = 5/2 + i*sqrt159/2

y = 5/2 - i*sqrt159/2

-for a = 5/2 and b = -i*sqrt159/2

x = 5/2 - i*sqrt159/2

y = 5/2 + i*sqrt159/2

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neela | High School Teacher | Valedictorian

Posted August 16, 2010 at 2:25 AM (Answer #2)

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x^4+y^4 = 257 and x+y = 5.

Solution:

(x+y)^4 =( x^4+y^4) +4xy(x^2+y^2) + 6(xy)^2

(x+y)^4 = (x^4+y^4) +4xy{(x+y)^2-2xy} +6xy

(x+y)^4 = (x^4+y^4) +4xy(x+y)^2 +(6-8)(xy)^2

5^4=257 +4(xy)(5)^2 -2(xy)^2. as x+y = 5 and x^4+y^4 = 257.

2(xy)^2 -100xy+625-257 = 0

2(xy)^2 -100xy+368 = 0

(xy)^2 -50xy +184 = o. This is a quadratic in xy.

(xy - 4)(xy -46 ) = 0

Therefore xy = 4 Or xy = 46.

xy = 4  and x+y = 5 gives: x-y = sqrt{(x+y)^2-4xy} =  sqrt(25-16) = 3.

So x +y+x-y = 2x = 5+3 =8. So x = 8/2 =4.

y = 5-x= 1.

So x= 4 and y =1.

xy =46 gives : x-y  = sqrt{(x+y)^2-4xy} = sqrt{25-4*46} = -159

x-y = sqrt(-159)

So x+y+x-y = 2x = 5+sqrt(-159),

 x = {5+sqrt(-159)}/2 and

 y = {5-sqrt(-159)}/2

 

 

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jeyaram | Student , Undergraduate | Valedictorian

Posted August 16, 2010 at 3:28 PM (Answer #3)

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x^4+y^4=257............(1)

x+y=5

x^4+y^4

=(x^2)^2+(y^2)^2

=(x^2+y^2)^2 - 2*x^2*y^2

={[(x+y)^2 - 2xy]^2 - 2*x^2*y^2}

={[(x+y)^2 - 2xy]^2 - 2(xy)^2}

=(25-2xy)^2 - 2(xy)^2                [ because (x+y)=5 ]

put z=xy

so          x^4+y^4=(25-2z)^2 - 2z^2

257=(25-2z)^2 - 2z^2         [ because (x^4+y^4)=257 ]

257=625-100z+4z^2- 2z^2

257=625-100z+2z^2

2z^2-100z+368=0

z^2-50z+184=0

z^2-4z-46z+184=0

z(z-4)-46(z-4)=0

(z-46)(z-4)=0

(z-46)=0         or            (z-4)=0

z=46               or                 z=4

so   xy=46       or                xy=4

 

when xy=46                            xy=4

(1)=> x=(5-y)                  (1)=> x=(5-y)

(5-y)y=46                           (5-y)y=4

y^2-5y+46=0               y^2-5y+4=0

Delta<0                     y^2-y-4y+4=0

then                         y(y-1)-4(y-1)=0

so                                 (y-4)(y-1)=0

.                             (y-4)=0    or     (y-1)=0

.                                   y=4    or          y=1

so                                x=1    or          x=4

.                   Answer is   {x=1;y=4}

.                                     {y=1;x=4}

 

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