Question

x+y+z=9 and xy+yz+zx=40 find x^2+y^2+z^2

Answer 1

X+y+z=9

Squaring both sides the avove eqn we get
X2+y2+z2+2(xy+yz+zx)=81
Substitute values of xy+yz+zx which is 40

So
X2+y2+z2=1

Hope you get your answer.

Answer 2

Answer: - The value of x² + y² + z² = 1.

Detailed solution: -

Given: -

Two equations are given.

x + y + z = 9.......equation(1)

And,

xy + yz + zx = 40...............equation(2)

We need to find the value of x² + y² + z².

• To find the value of the given equation x² + y² + z², we need to solve the sum using equation(1).
• We need to square both sides of the equation(1) and then put the value of equation(2) in it.

Therefore,

x + y + z = 9.......equation(1)

Squaring both sides of the given equation we get,

(x + y + z)² = 9²

x² + y² + z² + 2(xy + yz + zx) = $9*9$

Putting the value of equation(2), i.e., xy + yz + zx = 40, we get,

x² + y² + z² + $2*40$ = 81

x² + y² + z² + 80 = 81

(80 will be shifted to the right-hand side and will have subtraction sign.)

x² + y² + z² = 81 - 80

x² + y² + z² = 1

Therefore,

The value of x² + y² + z² = 1.

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