Question

What is the value of a-b where a2+b2-6a-6b+18=0

Answer 1

answer : a - b = 0

explanation : it is given that, a² + b² - 6a - 6b + 18 = 0

or, a² + b² - 6a - 6b + 9 + 9 = 0

or, (a² - 6a + 9) + (b² - 6b + 9) = 0

or, (a² - 2.a.3 + 3² ) + (a² - 2.b.3 + 3²) = 0

from algebraic identity,

• x² - 2xy + y² = (x - y)²

so, (a² - 2.a.3 + 3²) = (a - 3)²

and (a - 2.b. 3 + 3²) = (b - 3)²

now, (a² - 2.a.3 + 3² ) + (a² - 2.b.3 + 3²) = (a - 3)² + (b - 3)² = 0

LHS = RHS only if , (a - 3) = 0 and (b -3) = 0

hence, a = 3 and b = 3

now, we have to find value of (a - b)

so, a - b = 3 - 3 = 0