Answer:
An extraneous solution is a root of a transformed equation that is not a root of the original equation because it was excluded from the domain of the original equation. Example 1: Solve for x , 1x − 2+1x + 2=4(x − 2)(x + 2) .Set the equation to equal zero. (this ends up being √x+4−x+2=0 )
Plug this into the y= button on your TI-83/84 calculator.
Find the value of each of your solutions (go to 2nd->Calc->Value and enter your solution for x )
You should get zero as an answer for each of them.
Step-by-step explanation:
Extraneous solutions are not solutions at all. They arise from outside the problem, from the method of solution. They are extraneous because they are not solutions of the original problem. ... To tell if a "solution" is extraneous you need to go back to the original problem and check to see if it is actually a solution.
Answer:
This ans is easily solved by using algebraic identities a²- b² = (a+b) (a-b)
Step-by-step explanation:
1. convert into Identity
2. solve quadratic equation that formed.
the answer is C hope this helps
Answer:
16 < w < 21
Step-by-step explanation:
Given that :
Width if mirror = w
Length of mirror = l
L = 6 + w
Perimeter of mirror < 96 and > 76
Possible width of the mirror
Perimeter of a rectangle :
2(l + w)
2(6 + w + w)
= 2(6 + 2w)
= 12 + 4w
Hence,
76 < perimeter < 96
76 < 12 + 4w < 96
76 - 12 < 4w < 96 - 12
64 < 4w < 84
16 < w < 21
