Question

if x^2+y^2+z^2=69 and xy+yz+zx=50 find the value of x+y+z .​

Answer 1

Step-by-step explanation:

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

=69+2(50)

=69+100

=169

(x+y+z)2=169

x+y+z=13

Answer 2

$\huge{\mathcal{\green{Answer}}}$

$(x + y + z) {}^{2} = x {}^{2} + y {}^{2} + z {}^{2} + 2xy + 2yz + 2zx \\ (x + y + z) {}^{2} = 69 + 2(xy + yz + zx) \\ (x + y + z) {}^{2} = 69 + 2(50) \\ (x + y + z) {}^{2} = 69 + 100 \\ ( x + y + z) {}^{2} = 169 \\ x + y + z = \sqrt{169}$

$\huge{\mathcal{\orange{\boxed{x+y+z=13}}}}$