b^2-4b+4=16

Question

need help with quadratic equations

Answer 1

b² - 4b + 4 = 16 

 

Add -16 to both sides

b² - 4b + 4 - 16 = 16 - 16

b² - 4b -12 = 0

 

Factor the equation, you'll get

(b - 6)(b + 2) = 0

 

Equate both factors to 0

b - 6 = 0

b = 6

 

b + 2 = 0

b = - 2

 

Therefore, the two roots are (6, -2)

 

 

              

Answer 2

b²-4b+4 = 16

 

add -4 to both sides of the equation

 

b²-4b+4-4=16-4 , b²-4b=12  add -12 to both sides of the equation

 

b²-4b-12 = 12-12 , b²-4b -12 =0

 

factor equation ( b-6)(b+2) = 0 for this equation to be true either b-6 = 0 , or b+2=0

 

b-6 = 0 add +6 to both sides of the equation b-6+6=+6 or b=6

b+2 =0 add -2 to both sides of the equation  b+2-2 = -2 0r b =-2

 

The two roots are therefore ( 6,-2)

 

Proof:

 

Substitute b = 6 into b²-4b-12 = 0, (6)²-4(6) -12 = 0

Substitute b = -2 into b²-4b -12=0 , (-2)²-4(-2) -12 = 0

 

Therefore both roots are true