Question

Complete the square to solve 3x2 + 12x = 9.

+
ОА. x = 2+ 17
Ов. х = 3 + 6
Ос. x=-2+ 7
Op. x= -2+ 13

Answer 1

`3 {x}^(2) + 12x = 9 \\ 3 {x}^(2) + 12x + ( - 9) = 9 + ( - 9) \\ 3 {x}^(2) + 12x - 9 = 9 - 9 \\ 3 {x}^(2) + 12x - 9 = 0 \\ 3 ({x}^(2) + 4x - 3) = 0 \\ 3 ({x}^(2) + 4x - 3) * (1)/(3) = 0 * (1)/(3) \\ ({x}^(2) + 4x - 3)3 * (1)/(3) = 0 * (1)/(3) \\ ({x}^(2) + 4x - 3) * 1 = 0 \\ {x}^(2) + 4x - 3 \ = 0 \\ {x}^(2) + 4x - 3 + 3\ = 0 + 3 \\ {x}^(2) + 4x + 0\ = 3 \\ {x}^(2) + 4x \ =3 \\ {x}^(2) + 2(2)x + 2^(2) = 3 + {2}^(2) \\ {(x + 2)}^(2) = 3 + 4 \\ (x + 2) ^(2) = 7 \\ { ({(x + 2)}^(2)) }^{ (1)/(2) } = 7 ^{ (1)/(2) } \\ x + 2 = \ \sqrt[2]{ {7}^(1) } \\ x + 2 = \sqrt[2]{7} \\ x + 2 = √(7) \\ x + 2 + ( - 2) = √(7) + ( - 2) \\ x + 2 - 2 = ( - 2) + √(7) \\ x + 0 = - 2 + √(7) \\ x = - 2 + √(7)`

at √7 the values will be + and -

that is x = -2 + (+√7)

x = -2 + √7

and

x = -2 + (-√7)

x = -2 - √7

so x = -2±√7