Question

If (x – 3)2 + (y – 5)2 + (z – 4)2 = 0, then the value of x2/9+y2/25+z2/16 is.

Answer 1

HELLO DEAR,

HELLO DEAR,GIVEN THAT:-

(X - 3)² + (Y - 5)² + (Z - 4)² = 0

on soling,

we get,

=> (x² + 9 - 6x) + (y² + 25 - 10y) + (z² + 16 - 8z) = 0

on comparing both side we get quadratic equations

so, x² - 6x + 9 = 0--------------( 1 )

y² - 10y + 25 = 0-----------( 2)

z² - 8z + 16 = 0-------------( 3)

solving, --- ( 1 )

x² - 6x + 9 = 0

=> x² - 3x - 3x + 9 = 0

=> (x - 3)(x - 3) = 0

=> x = 3, x = 3

similarly,

solving---( 2 )

y² - 10y + 25 = 0

=> y² - 5y - 5y + 25 = 0

=> (y - 5)(y - 5) = 0

=> y = 5 , y = 5

similarly, in equation---- ( 3 )

z² - 8z + 16 = 0

=>(z - 4)(z - 4) = 0

=> z = 4 , z = 4

now,

we have to find:- x²/9 + y²/25 + z²/16

putting values of x,y,z in above eqaution

we get,

9/9 + 25/25 + 16/16 = 1 + 1 + 1 = 3

hence, x²/9 + y²/25 + z²/16 = 3

one more simple way we can do,

given equation is (X - 3)² + (Y - 5)² + (Z - 4)² = 0

on comparing both side we get,

(x - 3) = 0 => x = 3

(y - 5) = 0 => y = 5

(z - 4) = 0 => z = 4

therefore, x²/9 + y²/25 + z²/16 = 3

I HOPE IT'S HELP YOU DEAR,

THANKS