Question

if a^2+b^2+c^2 +170 = 2(8a+5b-9c) then find the value of (4a+8b-c)^0.5

Answer 1

Answer:

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Answer 2

SOLUTION

GIVEN

$\displaystyle \sf{ {a}^{2} + {b}^{2} + {c}^{2} = 2(8a + 5b - 9c) }$

TO DETERMINE

The value of

$\displaystyle \sf{ {(4a + 8b - c)}^{0.5} }$

EVALUATION

$\displaystyle \sf{ {a}^{2} + {b}^{2} + {c}^{2} = 2(8a + 5b - 9c) }$

$\displaystyle \sf{ \implies \: {a}^{2} + {b}^{2} + {c}^{2} - 2(8a + 5b - 9c) = 0 }$

$\displaystyle \sf{ \implies \: {a}^{2} - 16a + 64 + {b}^{2} - 10b + 25+ {c}^{2} + 18c + 81 = 0 }$

$\displaystyle \sf{ \implies \: {(a - 8)}^{2} + {(b - 5)}^{2} + {(c + 9)}^{2} = 0 }$

We know if the sum of the squares of three Real Numbers are zero then they are separately zero

Thus we get

a - 8 = 0 , b - 5 = 0 , c + 9 = 0

Consequently we get

a = 8 , b = 5 , c = - 9

$\displaystyle \sf{ {(4a + 8b - c)}^{0.5} }$

$\displaystyle \sf{ = {(32 + 40 + 9)}^{0.5} }$

$\displaystyle \sf{ = {(81)}^{0.5} }$

$\displaystyle \sf{ = {(81)}^{ \frac{1}{2} } }$

$= 9$

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