If (1)/(x) + (1)/(y) + (1)/(z) = 0 and {x}^(2) + {y}^(2) + {z}^(2) = 25 , then x + y + z will be? Z​

Question

If


`(1)/(x) + (1)/(y) + (1)/(z) = 0`
and

`{x}^(2) + {y}^(2) + {z}^(2) = 25`
, then x + y + z will be? Z​

Answer 1

• 5

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Explanation:

Given,

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`{ \longrightarrow \qquad{ { \sf{ (1)/(x) + (1)/(y) + (1)/(z) = 0}}}}`

⠀

`{ \longrightarrow \qquad{ { \sf{ {x}^(2) + {y}^(2) + {z}^(2) = 25 }}}}`

⠀

To Find :

`{ \longrightarrow \qquad{ { \sf{ x + y + z }}}}`

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Solution :

`{ \longrightarrow \qquad{ { \sf{ (1)/(x) + (1)/(y) + (1)/(z) = 0}}}}`

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`{ \longrightarrow \qquad{ { \sf{ (yz + zx + xy)/(xyz) = 0}}}}`

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`{ \longrightarrow \qquad{ { \sf{ {yz + zx + xy} = 0}}}}`

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Multiplying both sides by 2 :

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`{ \longrightarrow \qquad{ { \sf{ 2( {yz + zx + xy} )= 0.2}}}}`

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`{ \longrightarrow \qquad{ { \sf{ 2( {yz + zx + xy} )= 0}}}}`

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We know,

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

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As, It is given that,

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`{ \longrightarrow \qquad{ { \sf{ {x}^(2) + {y}^(2) + {z}^(2) = 25 }}}}`

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and as we get,

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`{ \longrightarrow \qquad{ { \sf{ 2( {yz + zx + xy} )= 0}}}} \\ \\{ \longrightarrow \qquad{ { \sf{ 2( {xy + yz + zx} )= 0}}}}`

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Now, We'll substitute it in the formula :

`{ \longrightarrow \qquad{ { \sf{ {(x + y + z)}^(2) = 25 + 0 }}}}`

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`{ \longrightarrow \qquad{ { \sf{ {(x + y + z)}^(2) = {25} }}}}`

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`{ \longrightarrow \qquad{ { \sf{ {x + y + z}^{} = √(25) }}}}`

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`{ \longrightarrow \qquad{ { \bf{ \pmb {x + y + z}^{} = {5} }}}}`

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Therefore,

• The value of x + y + z is 5 .

Answer 2

5

Explanation:

We would like to find out the value of
`x + y + z` from the given equations , which are ;

`\longrightarrow (1)/(x)+(1)/(y)+(1)/(z)=0`

`\longrightarrow \\ x^2+y^2+z^2=25`

Now consider ,

`\longrightarrow (1)/(x)+(1)/(y)+(1)/(z)=0\\`

`\longrightarrow (xy + yz + zx )/(xyz )=0\\`

`\longrightarrow xy + yz + zx = 0(xyz)\\`

`\longrightarrow xy + yz + zx = 0\\`

`\longrightarrow 2(xy + yz + zx)=0(2)\\`

`\longrightarrow 2(xy + yz + zx)=0`

Now recall the identity ,

`\longrightarrow (a + b + c)^2=a^2+b^2+c^2+2(ab + bc + ca)`

Plug in the values ,

`\longrightarrow (x+y+z)^2= 25 + 0\\`

`\longrightarrow (x + y + z )^2=25\\`

`\longrightarrow (x + y + z)=√(25)\\`

`\longrightarrow \underline{\underline{\boldsymbol{ x + y + z = 5}}}{}`

And we are done !