1-11/x+30/x^2=0
A.the solution set is ?
B.The solution set is the set of real numbers
or C. the solution set is an empty set
1 - 11/x + 30/x^2 = 0
times x^2
x^2 - 11x + 30 = 0
(x-5)(x-6) = 0
x = 5 or x = 6
To solve the given equation, we need to find the values of x that make the equation true.
Let's start by simplifying the equation by combining like terms and putting it in a single fraction:
(1 - 11/x) + (30/x^2) = 0
To add the fractions, we need a common denominator. Since the first fraction has x on the denominator, we multiply the second fraction by x/x to get:
(1 - 11/x) + (30x/x^2) = 0
Now, we have a common denominator:
(1 - 11/x + 30x) / x^2 = 0
To remove the denominator, we can multiply both sides of the equation by x^2:
1 - 11/x + 30x = 0
Next, let's rearrange the equation to isolate the term with the variable x:
30x - 11/x = -1
Now, we can multiply through by x to eliminate the fraction:
30x^2 - 11 = -x
Simplifying further:
30x^2 + x - 11 = 0
This equation is a quadratic equation. To find the solution set, we can use the quadratic formula. The quadratic formula is:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our quadratic equation of the form ax^2 + bx + c = 0, we have a = 30, b = 1, and c = -11. Substituting these values into the quadratic formula:
x = (-1 ± √(1^2 - 4(30)(-11))) / (2(30))
Simplifying further:
x = (-1 ± √(1 + 1320)) / 60
x = (-1 ± √1321) / 60
Therefore, the solution set is a set of real numbers (B), since the equation has real solutions.