Question

if x+y+z=10 and x²+y²+z²=40,then find xy+yz+zx
help me pls​

Answer 1

Answer:-

The value of xy+yz+zx is 30.

Given:

• x+y+z=10

• $\sf{x^{2}+y^{2}+z^{2}=40}$

To find:

• The value of xy+yz+zx.

Solution:-

According to the identity

$\sf{(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ac)}$

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

$\sf{\therefore{10^{2}=40+2(xy+yz+zx)}}$

$\sf{\therefore{100=40+2(xy+yz+zx)}}$

$\sf{\therefore{2(xy+yz+zx)=100-40}}$

$\sf{\therefore{2(xy+yz+zx)=60}}$

On dividing both sides by 2, we get

$\sf{xy+yz+zx=30}$

$\sf{\therefore}$ The value of xy+yz+zx is 30.

Answer 2

Given :

x + y + z = 10 and x² + y² + z² = 40

To find :

The value of (xy + yz + xz) ?

Solution :

$\sf Using \: identity \: \fbox{ {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ac) } \: , \: we \: get$

$</p><p>\Rightarrow \sf</p><p> {(10)}^{2} = 40 + 2(xy + yz + xz) \\ \\</p><p>\Rightarrow \sf</p><p> 100 = 40 + 2(xy + yz + xz) \\ \\</p><p>\Rightarrow \sf</p><p> 60 = 2(xy + yz + xz) \\ \\</p><p>\Rightarrow \sf</p><p>(xy + yz + xz) = \frac{60}{2} \\ \\</p><p>\Rightarrow \sf</p><p> (xy + yz + xz) = 30$

$\therefore \sf \underline{The \: value \: of \: (xy + yz + xz) \: is \: 30}$