Question

The roots of the quadratic equation 3(x+3)^2 = 48 are _______.

Answer 1

Answer:

x₁ = -7

x₂ = 1

Step-by-step explanation:

3(x+3)² = 48

3(x² + 6x + 9) = 48

x² + 6x + 9 = 48 / 3

x² + 6x + 9 = 16

x² + 6x + 9 - 16 = 0

x² + 6x - 7 = 0

(x + 7)(x - 1) = 0

x + 7 = 0

x₁ = -7

x - 1 = 0

x₂ = 1

Answer 2

Answer:

the zeroes(or roots) of the quadratic equation are 1 and -7$x^{2}$

Step-by-step explanation:

3(x+3)^2=48

($x^{2}$ + 6x + 9)=48/3

$x^{2}$ + 6x + 9=16

$x^{2}$ + 6x + 9-16=0

$x^{2}$ + 6x -7=0

By factorisation method,

$x^{2}$ -x + 7x -7 =0

x(x-1)+7(x-1)=0

(x-1)(x+7)=0

the zeroes are,

(x-1)=0 => x=1

and

(x+7)=0 =>x=-7

Therefore the zeroes(or roots) of the quadratic equation are 1 and -7