The equation 3x^2 - 30 = 3 can be rewritten as 3x^2 = 33. To solve this equation, we divide both sides by 3, giving us x^2 = 11. Taking the square root of both sides, we get x = ±√11. Therefore, there are two solutions to this equation.
On the other hand, the equation 3x^2 + 30 = 3 can be rewritten as 3x^2 = -27. Since squaring any real number will always give a non-negative result, there is no real number that satisfies this equation. Thus, there are no solutions to this equation.
To confirm this, we can try substituting different values for x into each equation:
For the equation 3x^2 - 30 = 3:
- If we substitute x = √11, we get 3(√11)^2 - 30 = 3(11) - 30 = 3, which is true.
- If we substitute x = -√11, we get 3(-√11)^2 - 30 = 3(11) - 30 = 3, which is also true.
For the equation 3x^2 + 30 = 3:
- If we substitute x = √11, we get 3(√11)^2 + 30 = 3(11) + 30 = 33 + 30 = 63, which is not equal to 3.
- If we substitute x = -√11, we get 3(-√11)^2 + 30 = 3(11) + 30 = 33 + 30 = 63, which is also not equal to 3.
These calculations confirm that the first equation has two solutions (√11 and -√11), while the second equation has no solution.
Why does the equation 3x^2 - 30 = 3 have two solutions but the equation 3x^2 + 30 = 3 has no solution?
Try some different values for x.
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