Question

If x+y+z=9 and xy+yz+zx = 24 , find the value of x2+y2+z2

Answer 1

Step-by-step explanation:

$\Huge\red{\mathfrak{ Answer }}$

x^2+y^2+z^2 = ?

Given

x+y+z= 9

xy+yz+xz= 24

we know that

(x+y+z)^2= x^2+y^2+z^2+2(xy+yz+xz)

= > (x+y+z)^2= (9)^2= 81

we have

81 = x^2+y^2+z^2+2(xy+yz+xz)

=> 81= x^2+y^2+z^2+2(24)

=> x^2+y^2+z^2= 81-48

=>x^2+y^2+z^2= 33

hope it helps ✌

Answer 2

Given:

x + y + z = 9

xy + yz + zx = 24

To Find:

The value of x² + y² + z².

Solution:

- We know that the formula is given as:

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

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

⇒ 9² = x² + y² + z² + 2 × 24

⇒ 81 = x² + y² + z² + 48

⇒ x² + y² + z² = 81 - 48

⇒ x² + y² + z² = 33

- So, the value of term (x² + y² + z²) is calculated as 33.