Question

if x^2+y^2+z^2=40 and xy+yz+zx=30, find x+y+z​

Answer 1

${\huge{\rm{\underline{\underline{Answer\checkmark}}}}}$

$\rm{Given→} \begin{cases}\sf{x^2 + y^2 + z^2=40}\\ \sf{xy+yz+zx=30}\end{cases}$

${\underline{\bold{\large{\bf{\it{To\;Find:-}}}}}}$ $\rm{\rightarrow\; x+y+z}$

Solution :–

Squaring x+y+z

$\mapsto \rm \: {(x + y + z)}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2(xy + yz + zx)$

We have the values of -

• x² + y² + z² = 40
• xy + yz + zx = 30

so , put these values ..

$\mapsto \rm \: {(x + y + z)}^{2} = 40 + 2(30)$

$\mapsto \rm \: {(x + y + z)}^{2} = 40 + 60$

$\mapsto \rm \: {(x + y + z)}^{2} = 100$

Take square root both sides

$\mapsto \rm \: \sqrt{ {(x + y + z)}^{2} } = \sqrt{100}$

$\mapsto \underline{ \boxed{ \sf{ \pink{x + y + z = 10}}}}$

So ,

The Final Answer is 10 .

_________________________

★ Some Important Formulas :–

• (a+b)² = a² + b² + 2ab

• (a-b)² = a² + b² - 2ab

• (-a+b)² = a² + b² - 2ab

• (a+b+c)² = a² + b² + c² + 2( ab + bc + ac )

• (a-b-c)² = a² + b² + c² + 2(-ab + bc - ac)

• (a+b)³ = a³ + b³ + 3ab(a+b)

• (a-b)³ = a³ - b³ - 3ab(a-b)

Answer 2

Answer:

Answer :

Given by : x^2+y^2+z^2=40,. XY+yz+zx =30

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

=(x^2+y^2+z^2)+2(XY+yz+zx)

= 40+2×30

=100

x+y+z=√100= 10