if q = ( x – 5 ) ^ 2 + ( x + 1 ) ^ 2 – 6 , then y is least when x =
let us transform the formula : q = ( x – 5 ) ² + ( x + 1 ) ² – 6 = x ² – 10 x + 25 + x ² + 2 x + 1 – 6 = 2 x ² – 8 x + 20 = 2 × ( x ² – 4 x + 10 ) = 2 × ( ( x ² – 4 x + 4 ) + 6 ) = 2 × ( ( x – 2 ) ² + 6 ) any square is greater or equal 0 . therefore the formula possess the least value when ( x – 2 ) ² = 0 . x – 2 = 0 x = 2 the correct answer is choice ( d ) .
negate(divide(divide(subtract(2, multiply(5, 2)), 2), 2))