Respuesta :

RecklessChicken

Please Follow through my step-by-step tutorial till the end:

1.) First, move the number(s) to the other side. (algebra rules)

2(x-3)²-32+32=0+32

2(x-3)² = 32

2.) Next, divide every term by 2 to factor out the 2.

2(x-3)²/2 = 32/2

(x-3)² = 16

3.) Third, square root both sides (algebraic rules)

[tex]\sqrt{(x-3)^2} = \sqrt{16}\\[/tex]

[tex](x-3) = \left \{ {{+4} \atop {-4}} \right.[/tex]

4.) Since the square root of 16 is both +4 and -4, we will have two answers. Next we split the equation in two forms.

(x-3) = 4                (x-3) = -4

x-3+3 = 4+3          x-3+3 = -4+3

x=7                        x=-1

5.) So we get two answers x= 7 and x= -1

szimbitskyy

Answer: x=7 and x= -1

Step by step explanation:

First, you divide every term by 2 to factor out the 2. Then you will add both sides by 16. Next, you will square root each side. Square root of a whole number is both positive and negative so we will be having two answers. We split the equation in two and solve it algebraically.

[tex]\frac{2(x-3)^2}{2} -\frac{32}{2} =\frac{0}{2} \\(x-3)^2 -16+16=0+16\\(x-3)^2 = 16\\\sqrt{(x-3)^2} = \sqrt{16} \\(x-3)= 4        :;:(x-3)=-4\\x=4+3   ::::::: x=-4+3\\       x=7        ::::::::::::::       x=-1[/tex]

Otras preguntas