I’m sorry if this is wrong, but wouldn’t it be 30;60? Again if I’m wrong someone else feel free to answer and correct me
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:
2 square root of 2 and -2 square root of 2
Step-by-step explanation:
Because x,y,z form a geometric sequence, therefore the common ratio is
r = y/x = z/y
That is,
y² = xz (1)
We are given:
x + y + z = 18
Therefore
x + z = 18 - y (2)
Also,
x² + y² + z² = 612
Therefore, from (1), obtain
x² + z² + xz = 612
(x + z)² - xz = 612
From (1) and (2), obtain
(18 - y)² - y² = 612
324 - 36y + y² - y² = 612
-36y = 288
y = -8
Answer: y = -8
