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.
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
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.
To determine if the equation √2x + 3 + 6 = 0 has a real solution or not, you need to solve the equation and see if the solution satisfies the given conditions.
Here's how you can solve the equation step by step:
1. Start by subtracting 6 from both sides of the equation:
√2x + 3 = -6
2. Next, subtract 3 from both sides:
√2x = -9
3. To eliminate the square root, square both sides of the equation:
(√2x)^2 = (-9)^2
2x = 81
4. Divide both sides of the equation by 2 to solve for x:
x = 81/2
Now, to determine if this solution satisfies the given equation (√2x + 3 + 6 = 0), substitute the value of x back into the equation and evaluate it:
√2(81/2) + 3 + 6 = 0
√162 + 3 + 6 = 0
12.728 + 3 + 6 = 0
21.728 ≠ 0
Since the equation does not hold true when x = 81/2, there is no real solution. Therefore, the correct answer choice for your practice test would be "No Real Solution."