Algebra

Algebra:

1.) Consider the following equation 

x² + 4y² + 9z² = 14x + 28y + 42z - 147 

where x, y and z are real numbers. Then the value of x + 2y + 3z is :

a.) 7

b.) 14

c.) 21

d.) not unique

Ans. : (c) 21 .

Solution:

x² + 4y² + 9z² = 14x + 28y + 42z - 147 

x² + 4y² + 9z² - 14x - 28y - 42z + 147 = 0

x² + 4y² + 9z² - 14x - 28y - 42z + 49 + 49 + 49 = 0

x² - 14x + 49 + 4y² - 28y + 49 + 9z² - 42z + 49 = 0

( x² - 14x + 49 ) + ( 4y² - 28y + 49 ) + ( 9z² - 42z + 49 ) = 0

( x - 7 )² + ( 2y - 7 )² + ( 3z - 7 )² = 0

Thus, each term must be separately zero.

( x - 7 )² = 0 

Take square root on both sides,

( x - 7 ) = 0

x - 7 = 0

x = 0 + 7

x = 7

( 2y - 7 )² = 0

Take square root on both sides,

( 2y - 7 ) = 0

2y - 7 = 0

2y = 0 + 7

2y = 7

y = 7/2

( 3z - 7 )² = 0

Take square root on both sides,

( 3z - 7 ) = 0

3z - 7 = 0

3z = 0 + 7

3z = 7

z = 7/3

x + 2y + 3z = 7 + 2 ( 7/2 ) + 3 ( 7/3 ) = 7 + 7 + 7 = 21 .

Then, the value of x + 2y + 3z is 21 .