Question

x+y+z=12 xy+yz+zx=37 find value of x,y,z
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Answer 1

Given:  

$x^2$ + $y^2$ + $z^2$ = 70,  x + y + z = 12 and xy + yz + zx = 37

We have to find, the values of x, y and z are:

Solution:

Using the algebraic identity:

$(x + y + z)^2$ = $x^2$ + $y^2$ + $z^2$  + 2(xy + yz + zx)

⇒ 12² = 70 + 2(37)

⇒ 144 = 70 + 74

⇒ 144 = 144

Thus, 3rd equation can be found if 2 equations are given.

We can not solve 3 variables with 2  Equations

There can be many possible solutions  

few are below :

x =  $\sqrt{37}$i, y = - $\sqrt{37}$i and z = 12

x = - $\sqrt{37}$i , y = $\sqrt{37}$i and z = 12

z = $\sqrt{37}$i,  y = - $\sqrt{37}$i and x = 12

Verification:  

$x^2$ + $y^2$ + $z^2$ =  - 37 - 37 + 144 = 70   [ ∵ $i^2$ = - 1]

x + y + z = $\sqrt{37}$i - $\sqrt{37}$i  + 12 = 12 and

xy + yz + zx = ($\sqrt{37}$i) (-$\sqrt{37}$i) + ($\sqrt{37}$i) 12  +  (-$\sqrt{37}$i) 12  

= - 37$i^2$ = 37

Not Enough Details to find Unique Solutions

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