1. If `x+y+z=9` and `x^2+y^2+z^2=29`, then find `xy+yz+zx`
Solution:
Here `X+Y+Z=9` and `X^2+Y^2+Z^2=29`
Now, We know that
`(x+y+z)^2=(x^2+y^2+z^2)+2(xy+yz+zx)`
`:.2(xy+yz+zx)=(x+y+z)^2-(x^2+y^2+z^2)`
`:.2(xy+yz+zx)=9^2-29`
`:.2(xy+yz+zx)=81-29`
`:.2(xy+yz+zx)=52`
`:.(xy+yz+zx)=52/2`
`:.(xy+yz+zx)=26`
2. If `x^2+y^2+z^2=29` and `xy+yz+zx=-14`, then find `x+y+z`
Solution:
Here `xy+yz+zx=-14` and `x^2+y^2+z^2=29`
Now, We know that
`(x+y+z)^2=(x^2+y^2+z^2)+2(xy+yz+zx)`
`:.(x+y+z)^2=29+2*(-14)`
`:.(x+y+z)^2=29-28`
`:.(x+y+z)^2=1`
`:.(x+y+z)=1`
3. If `x+y+z=1` and `x^2+y^2+z^2=29`, then find `xy+yz+zx`
Solution:
Here `X+Y+Z=1` and `X^2+Y^2+Z^2=29`
Now, We know that
`(x+y+z)^2=(x^2+y^2+z^2)+2(xy+yz+zx)`
`:.2(xy+yz+zx)=(x+y+z)^2-(x^2+y^2+z^2)`
`:.2(xy+yz+zx)=1^2-29`
`:.2(xy+yz+zx)=1-29`
`:.2(xy+yz+zx)=-28`
`:.(xy+yz+zx)=-28/2`
`:.(xy+yz+zx)=-14`
This material is intended as a summary. Use your textbook for detail explanation.
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