x^2-6x+12=19

Question

how do you complete the square to this function i have a math test tomarow and i do not know or understand this please help 

 

Answer 1

x² -6x + 12 = 19

==> x² -6x + 12 -12 = 19 -12   ------------ subtracting 12 on both sides

==> x² -6x = 7

==> x² - 2 * x * 3  = 7 

==> x² - 2* x * 3 + 3² = 7 + 3^{2  ------- to get perfect square adding} 3²  on both sides [a²-2ab+b² = (a-b)²]

==> (x - 3)² = 7 + 9   ------------- using formula a² - 2ab + b² = (a-b)²

==> (x-3)² = 16

==>√(x-3)² = √16 ---------------- square rooting on both sides 

==> x-3 = ±4

==>i) If  x-3 = 4 -->x = 7 -------- adding both sides with 3

==> ii) If x-3 = -4 --> x = -1 ------------ adding both sides with 3

 

Answer 2

x^2 - 6x + 12 = 19           1. Move the 12 over by subtracting

x^2 -6x = 19 -12              2. simplify

x^2 -6x = 7    2.               3. take the middle term (-6x)  work with number only & divide by 2 & square it  

x^2 -6x + (-6/2)^2 = 7 + (-6/2)^2    4. on both side of the equation so this does not change the given eq.

x^2 -6x + 9 = 7 + 9         5. you can skip this step if you like but this helps understand 

(x - 3) (x - 3) = 16           6. factor the left side  & simplify the right

(x - 3)^2 = 16                7. Rewrite it

Completing the square is done.  Solve only if told. just as Ralph has done.

Answer 3

x² - 6x + 12 = 19

x² -6x = 19 -12   --> x² -6x = 7

(x - 3)² = 7 + 9

(x - 3)² = 16

√(x - 3)² = ±√16

x - 3 = 4

x = 4 + 3   --> x = 7

x - 3 = -4

x = -4 + 3 --> x = -1