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.
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.