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

1.	If a^2+b^2+c^2 +170 = 2(8a+5b-9c) then find the value of (4a+8b-c)^0.5
Answer» 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 $\displaystyle \sf{ {a}^{2} + {b}^{2} + {c}^{2} = 2(8a + 5b - 9c) }$ $\displaystyle \sf{ \implies \: {a}^{2} + {b}^{2} + {c}^{2} - 2(8a + 5b - 9c) = 0 }$ $\displaystyle \sf{ \implies \: {a}^{2} - 16a + 64 + {b}^{2} - 10b + 25+ {c}^{2} + 18c + 81 = 0 }$ $\displaystyle \sf{ \implies \: {(a - <klux>8</klux>)}^{2} + {(b - 5)}^{2} + {(c + 9)}^{2} = 0 }$ 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 $\displaystyle \sf{ {(4a + 8b - c)}^{0.5} }$ $\displaystyle \sf{ = {(32 + 40 + 9)}^{0.5} }$ $\displaystyle \sf{ = {(81)}^{0.5} }$ $\displaystyle \sf{ = {(81)}^{ \frac{1}{2} } }$ $= 9$ ━━━━━━━━━━━━━━━━ Learn more from Brainly :- 1. solve: 49³-30³+(....)³= 3 × 49 × 30 × 19 brainly.in/question/17577612 2. If x+1/x=3 then, x^7+1/x^7= brainly.in/question/23295932

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