To solve the equation 3x^2 - 30 = 3, we can rearrange it as follows:
3x^2 - 33 = 0
The equation can be factored as:
3(x^2 - 11) = 0
This equation has two potential solutions:
x^2 - 11 = 0
(x - √11)(x + √11) = 0
So, the solutions are x = √11 and x = -√11.
On the other hand, let's solve the equation 3x^2 + 30 = 3:
3x^2 + 27 = 0
This equation cannot be factored with real numbers. If we solve it using the quadratic formula, we get:
x = (-b ± √(b^2 - 4ac)) / 2a
For this equation, a = 3, b = 0, and c = 27. Plugging these values into the quadratic formula, we get:
x = (0 ± √(0^2 - 4*3*27)) / (2*3)
x = ±√(-324) / 6
Since the square root of a negative number is not defined in the real number system, there are no real solutions to this equation.
Trying different values for x can help illustrate this:
For the equation 3x^2 - 30 = 3:
- For x = 0, we get 3(0)^2 - 30 = 3, which is not true.
- For x = 1, we get 3(1)^2 - 30 = 3 - 30 = -27, which is not true.
- For x = -1, we get 3(-1)^2 - 30 = 3 - 30 = -27, which is not true.
On the other hand, for the equation 3x^2 + 30 = 3:
- For x = 0, we get 3(0)^2 + 30 = 0 + 30 = 30, which is not true.
- For x = 1, we get 3(1)^2 + 30 = 3 + 30 = 33, which is not true.
- For x = -1, we get 3(-1)^2 + 30 = 3 + 30 = 33, which is not true.
In both cases, none of the values tested satisfies the equation. Thus, the equations have no real solutions or solutions in the real number system.
Why does the equation 3x2 - 30 = 3 have two solutions but the equation 3x2 + 30 = 3 has no solution?
Try some different values for x.
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