NTSE Question

Algebra Level 2

If a , b , c a, b, c are real numbers such that a 2 + b 2 + 2 c 2 − 4 a + 2 c − 2 b c + 5 = 0 , a^2 + b^2 +2c^2 - 4a+2c -2bc +5 = 0 , then what is the value of a + b − c a+b-c ?


The answer is 2.

View wiki
Alamuru Ganesh by Alamuru Ganesh , 21, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

We can factor the left hand side into ( a − 2 ) 2 + ( c − b ) 2 + ( c + 1 ) 2 = 0. (a - 2)^2 +(c - b)^2 +(c + 1)^2 =0.

Because the left hand side consists of a sum of squares and the right hand side is 0, it must be that each square is 0. This implies that a = 2 a = 2 , that b = c b = c , and that c = − 1 c = -1 .

Therefore, a + b − c = 2 − 1 + 1 = 2 . a + b - c = 2 - 1 + 1 = \boxed{2}.

34 Helpful 1 Interesting 3 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...