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 find the solutions to an equation, we need to solve for the variable x.
For the equation 3x^2 - 30 = 3:
Step 1: Add 30 to both sides of the equation.
3x^2 - 30 + 30 = 3 + 30
3x^2 = 33
Step 2: Divide both sides of the equation by 3.
(3x^2)/3 = 33/3
x^2 = 11
Step 3: Take the square root of both sides of the equation.
√(x^2) = √11
x = ±√11
Hence, this equation has two solutions: x = √11 and x = -√11.
Now, let's try the equation 3x^2 + 30 = 3:
Step 1: Subtract 30 from both sides of the equation.
3x^2 + 30 - 30 = 3 - 30
3x^2 = -27
Step 2: Divide both sides of the equation by 3.
(3x^2)/3 = -27/3
x^2 = -9
Taking the square root of a negative number results in an imaginary number. Since this equation has an imaginary solution (x = ±3i), it does not have a real solution.
By substituting different values for x, you can also observe that the equation 3x^2 - 30 = 3 has two solutions. For example, when x = 1, the equation becomes 3(1^2) - 30 = 3, which is true. Similarly, when x = -1, the equation becomes 3(-1^2) - 30 = 3, which is also true. On the other hand, for the equation 3x^2 + 30 = 3, when you substitute different values for x, none of them will satisfy the equation.