if you write it as
√(2x) = -9
you can see it has no real solutions, since √z is never negative.
Recall that even though
(2)^2 = 4 and (-2)^2 = 4, √4 is 2, not ±2
Hello! I am trying to write a study guide for an upcoming test, because my class does not provide one. This is the practice test question. I know the answer to this equation; however, I wish to know HOW to solve it for the real test.
The equation:
√2x + 3 + 6 = 0
Answer choices:
-No Real Solution (The answer is this one)
-One Real Solution
-Infinitely Many Real Solutions
My main question is: How would I know if it's a "No Real Solution" or not? Any help is much appreciated! Thank you in advance.
3 answers
Got it, thank you so much for your help oobleck, I really appreciate it!
sqrt(2x) + 3 + 6 = 0.
sqrt(2x) = -9,
2x =(-9)^2 = 81,
X = 40.5.
Check: sqrt(2*40.5) +3 +6 = 0.).
sqrt(81) + 3 + 6 = 0,
-9 + 9 = 0.
Note: The sq. root of 81 = +9 or (-9). I chose (-9), because it satisfies the Eq.
If these operations are legal, we have one solution.
I hope to get feedback from other tutors.
sqrt(2x) = -9,
2x =(-9)^2 = 81,
X = 40.5.
Check: sqrt(2*40.5) +3 +6 = 0.).
sqrt(81) + 3 + 6 = 0,
-9 + 9 = 0.
Note: The sq. root of 81 = +9 or (-9). I chose (-9), because it satisfies the Eq.
If these operations are legal, we have one solution.
I hope to get feedback from other tutors.
1 answer