Question

x+y+z=12, x² + y² + z² =70, xy+yz+zx=37, then x=?, y=?, z=?​

Answer 1

•Given:- x+ y+ z = 12 ,xy + yz + zx= 37 Xsquare + y square+ z square = 70

• To find:- the value of x,y,z

•Solution:-

According to the question,

(x + y + z)² =  x² + y² + z² + 2(xy + yz + zx)

or,12² = 70 + 2(37)

or,144 = 70 + 74

or, 144 = 144

as there are 3 equations and we will get

x =  √37i    y = -√37 i     z = 12.....(1)

x =  -√37i    y =  √37 i     z = 12 .....(2)

x = 12  y = -√37 i z=√37i......(3)

x = 12.  y =  √37 i     z= -√37i....(4)

x =  √37i  y = 12  z = -√37 i ....(5)   

x =  -√37i   y = 12.  z =  √37 i  ....(6)  

if we put any kind of value of x, z and y from above

So we can get...

according to (1)

x² + y² + z² =  -37 - 37 + 144 = 70   (i² = -1)

or,x + y + z = √37 i -√37 i  + 12 = 12

or,xy + yz + zx = (√37i) × (-√37 i) + (-√37i) 12 + (√37i) 12 = 37

according to (2)

x² + y² + z² =  -37 - 37 + 144 = 70   (i² = -1)

or,x + y + z = -√37 i +√37 i  + 12 = 12

or,xy + yz + zx = (-√37i) × (√37 i) + (√37i) 12 + (-√37i) 12 = 37

according to (3)

x² + y² + z² =144-37-37 = 70   (i² = -1)

or,x + y + z = 12-√37i+√37i = 12

or,xy + yz + zx =  (-√37i) 12+(√37i) × (-√37 i)+ (√37i) 12 = 37

according to (4)

x² + y² + z² =144-37-37 = 70   (i² = -1)

or,x + y + z = 12+√37i-√37i = 12

or,xy + yz + zx =  (√37i) 12+(√37i) × (-√37 i)+ (-√37i) 12 = 37

according to (5)

x² + y² + z² =  -37 + 144 - 37= 70   (i² = -1)

or,x + y + z = √37 i +12 -√37 i = 12

or,xy + yz + zx = (√37i) 12 +(-√37i) 12 + (-√37i) × (√37 i)= 37

according to (6)

x² + y² + z² =  -37 + 144 - 37= 70   (i² = -1)

or,x + y + z = -√37 i +12 +√37 i = 12

or,xy + yz + zx = (-√37i) 12 +(√37i) 12 + (-√37i) × (√37 i)= 37

Answer 2

Answer:

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