Answer:
1 and 4 has same pairs of solution.
Step-by-step explanation:
Solutions for the given equations are:
\begin{gathered}1. (x+6)(x - 6) = 0\\(x+6) = 0, (x-6) = 0\\x = -6, x = 6\\\\2. (x + 6)(x + 6) = 0\\(x+6) = 0, (x+6) = 0\\x = -6, x = -6\\\\3. (x - 6)(x - 6) = 0\\(x-6) = 0, (x-6) = 0\\x = 6, x = 6\\\\4. (2x + 12)(2x - 12) = 0\\(2x+12) = 0, (2x-12) = 0\\x = -6, x = 6\\\\5. (2x - 12)(x - 12) = 0\\(2x-12) = 0, (x-12) = 0\\x = 6, x = 12\\\\6. (x+12)(x - 12) = 0\\(x+12) = 0, (x-12) = 0\\x = -12, x = 12\\\\7. (x +12)(x-6) = 0\\(x+12) = 0, (x-6) = 0\\x = -12, x = 6\end{gathered}
1.(x+6)(x−6)=0
(x+6)=0,(x−6)=0
x=−6,x=6
2.(x+6)(x+6)=0
(x+6)=0,(x+6)=0
x=−6,x=−6
3.(x−6)(x−6)=0
(x−6)=0,(x−6)=0
x=6,x=6
4.(2x+12)(2x−12)=0
(2x+12)=0,(2x−12)=0
x=−6,x=6
5.(2x−12)(x−12)=0
(2x−12)=0,(x−12)=0
x=6,x=12
6.(x+12)(x−12)=0
(x+12)=0,(x−12)=0
x=−12,x=12
7.(x+12)(x−6)=0
(x+12)=0,(x−6)=0
x=−12,x=6
(x+6)(x - 6) = 0 and (2x + 12)(2x - 12) = 0 have same pair of solutions.
Step-by-step explanation:
there you go :)
