Question

prove that
.......
please
meri notebook mil nahi rahi​

Answer 1

$\bold{Given }= > if \\ {x}^{2} + 4 {y}^{2} + 25 {z}^{2} = 2 xy + 10yz + 5zx$

$\bold{To \: prove} = > x = 2y = 5z$

$concept \: used = > \\ 1) {(x - y )}^{2} = {x}^{2} + {y}^{2} - 2x \: y$

$2) \: value \: of \: square \: can \: not \: be \: negative \:$

$3)if \: sum \: of \: square \: quantities \: is \: equal \: to \: zero \\ if \: quantities \: individullay \: equal \: to \: zero$

$\bold{Solution} = > \\ {x}^{2} + 4 {y}^{2} + 25 {z}^{2} = 2xy + 10yz + 5zx$

$= > {(x)}^{2} + {(2y)}^{2} + {(5z)}^{2} = (x) \: (2y) + (2y) \: (5z) + (x) \: (5z)$

$let \: x = a \: and \: 2y = b \: and \: 5z = c$

$= > {a}^{2} + {b}^{2} + {c}^{2} = ab \: + bc \: + ca$

$multiplying \: whole \: equation \: by \: 2$

$= > 2 {a}^{2} + 2 {b}^{2} + 2 {c}^{2} = 2ab \: + \: 2bc \: + 2ca$

$= > 2 {a}^{2} + 2 {b}^{2} + 2 {c}^{2} - 2ab - 2bc - 2ca = 0$

$= > ( {a}^{2} + {b}^{2} - 2ab) + ( {b}^{2} + {c}^{2} - 2bc) + ( {c}^{2} + {a}^{2} - 2ca) = 0$

$= > {(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2} = 0$

$sum \: of \: square \: quantities \: is \: equal\: to \: zero \: if \: only \\ a - b = 0 \: = > a = b \\ and \: if \\ b - c = 0 = > b = c \\ and \: if \\ a - c = 0 = > a = c \\ so \: a = b = c \\ putting \: a = x \: and \: b = 2y \: and \:c = 5z \\ = > x = 2y = 5z$