Three real numbers x, y, z are such that x2 + 6y = 􀀀17, y2 + 4z = 1 and z2 + 2x = 2. What is the value of x2 + y2 + z2?9:01 PM

x^2 + 6y = 17   

y^2 + 4z = 1    

z^2 + 2x = 2  

WolframAlpha shows a couple of real solutions.......

I leave the rest for you..........

Here's the worked out solution.........

x^2 + 6y = 17    →     y = [17 - x^2] / 6 

y^2 + 4z = 1    →       z = [ 1  - y^2 ] / 4 →   z =  [ 1 -  [(17 - x^2)/6]^2]^2/4

z^2 + 2x = 2  →        [ 1 - (1 - [(17 - x^2)/6]^2)]^2 + 2x  = 2       (3)

Using  (3).....we have

  [ 1 -  [(17 - x^2)/6]^2)]^2 / 4 + 2x  = 2

[ 1 - [ 289 - 34x^2 + x^4]/ 36]^2 /16  + 2x = 2   simplify

[(1/36) (-x^4 + 34x^2 - 289) + 1]^2 / 16  + 2x = 2

[ 1 / 20736] [ (-x^4 + 34x^2 - 289) + 36]^2 + 2x = 2

[ -x^4 + 34x^2 - 253]^2 / 20736  + 2x  = 2

[x^8 -68x^6 +1662x^4-17204x^2 + 64009] / 20736  + 2x  = 2

[x^8 -68x^6 +1662x^4-17204x^2 + 41472x +  64009]  / 20736   = 2

[x^8 -68x^6 +1662x^4-17204x^2 + 41472x +  64009]  = 41472

x^8 -68x^6 + 1662x^4 - 17204x^2 + 41472x + 64009 - 41472 = 0

x^8 -68x^6 + 1662x^4 - 17204x^2 + 41472x + 22537 = 0

Kes, this will be too difficult to solve algebraically........we can either find the real solutions by a graph or by using a solver........WolframAlpha gives the real solutions of :

x ≈ -6.42773      and     x = -0.458115

So  x^2   = either ≈  41.3157129529    or   ≈  0.209869353225

And     x^2 +  6y  = 17   ....so........

41.3157129529 + 6y = 17   →  y ≈ -4.0526   → y^2 ≈  16.42356676     .......or......

0.209869353225 + 6y = 17  →  y ≈ 2.798   → y^2 ≈  7.829

And y^2 + 4z = 1    →   16.42356676 + 4z = 1  → z  ≈ 3.856 → z^2  ≈ 14.869    .....or......

 7.829 + 4z  = 1  →  z ≈ -1.707     →  z^2 ≈  2.914

So...in one case   x^2 + y^2 + z^2  =    41.3157129529 +   16.42356676  +  14.869  = 72.6082797129

OR

x^2 + y^2 + z^2  =    0.209869353225 + 7.829  + 2.914  = 10.952869353225

x2+6y=-17  .....................             (1)

y2+4z=1     .....................             (2)

z2+2x=2     ......................            (3)

now add 1,2&3

we get x2+6y+y2+4z+z2+2x=-17+1+2

                                          =-14

we can rewrite the equation as (x+1)^2+(y+3)^2+(z+2)^2=-14+1+4+9

                                                                                       =0

we did the above step by adding (1^2=1)+(2^2=4)+(3^2=9) on both side of the equation and factorizing the LHS.

so the equation becomes (x+1)^2+(y+3)^2+(z+2)^2=0

therefore to get 0 each of the three squares should be equal to 0

(x+1)^2=0              x=-1

(y+3)^2=0              y=-3

(z+2)^2=0              z=-2

so the value of x2+y2+z2=(-1)^2+(-3)^2+(-2)^2

                                      =1+4+9=14

ans-14

Guest # 3

Did you check your "solutions" by substituting to see if they balance? Well, they don't!!.

x2+6y=-17  .....................             (1)

y2+4z=1     .....................             (2)

z2+2x=2     ......................            (3)

now add 1,2&3

we get x2+6y+y2+4z+z2+2x=-17+1+2

                                          =-14

adding 1,4,9 on LHS and RHS

i.e.

x^2 +6y+y^2+4z+z^2+2x++9+4+1= -14+1+9+4

therefore

(x^2+2x+1)+(y^2+6y+9)+(z^2+4z+4)=0

(x+1)^2+(y+3)^2+(z+2)^2=0

we did the above step by adding (1^2=1)+(2^2=4)+(3^2=9) on both side of the equation and factorizing the LHS.

so the equation becomes (x+1)^2+(y+3)^2+(z+2)^2=0

therefore to get 0 each of the three squares should be equal to 0

(x+1)^2=0              x=-1

(y+3)^2=0              y=-3

(z+2)^2=0              z=-2

so the value of x2+y2+z2=(-1)^2+(-3)^2+(-2)^2

                                      =1+4+9=14

ans-14

#guest4

yes i have checked the solution by using substituition method.       and if still you are not satisfied by my solution then plz be clear with your question so that even i can be more clear in my explaination..!!!