Respuesta :
Assuming the given equation is (x-2)^2 = 9, Edna is partially correct. Saying x-2 = 3 is halfway there in terms of getting the ful solution. She forgot about the minus form of the plus minus. If (x-2)^2 = 9, then applying the square root to both sides leads to these two equations: x-2 = 3 or x-2 = -3. So we have a plus 3 and then a minus 3.
Answer:x=5,-1
Step-by-step explanation:
Edna says that when \left ( x-2\right )^2=9[/tex]
then x-2=3
such that x=5
but this is not true as when we put the value of x in equation then it will not satisfy the equation.
It can be solved by taking the terms either on LHS or RHS
[tex]\left ( x-2\right )^2-9=0[/tex]
[tex]x-2=\pm 3[/tex]
x=5
x=-1
