x+5 1
------- - 1= -------
x^2-2x x^2-2x
To solve this equation, we need to find the value of x that satisfies the equation. Let's go step by step.
First, we need to simplify the expression on both sides of the equation:
The left side can be simplified by combining the fractions:
(x + 5)/(x^2 - 2x) - 1 = (x + 5 - (x^2 - 2x))/(x^2 - 2x) = (-x^2 + 2x + 5)/(x^2 - 2x)
The right side remains the same:
1/(x^2 - 2x)
Now we have the equation:
(-x^2 + 2x + 5)/(x^2 - 2x) = 1/(x^2 - 2x)
To proceed, we can cross multiply:
(-x^2 + 2x + 5)(x^2 - 2x) = (x^2 - 2x)(1)
Simplify both sides of the equation:
-x^4 + 2x^3 + 5x^2 - 2x^3 + 4x^2 - 10x = x^2 - 2x
Combine like terms:
-x^4 + 3x^3 + 9x^2 - 12x = x^2 - 2x
Move all terms to one side of the equation:
-x^4 + 3x^3 + 8x^2 - 10x = 0
Next, let's factor out the common terms:
-x(x^3 - 3x^2 - 8x + 10) = 0
Now, we have the equation:
-x(x - 1)(x^2 - 2x + 10) = 0
To find the value of x, we set each factor equal to zero:
1) x = 0
2) x - 1 = 0, which gives x = 1
3) x^2 - 2x + 10 = 0 - This is a quadratic equation.
Now, we can solve the quadratic equation. We can either use the quadratic formula or complete the square. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our quadratic equation, a = 1, b = -2, and c = 10.
x = (-(-2) ± √((-2)^2 - 4(1)(10))) / (2(1))
x = (2 ± √(4 - 40)) / 2
x = (2 ± √(-36)) / 2
Since the square root of a negative number is not a real number, we have no real solutions for the quadratic equation.
Therefore, the solution to the given equation is:
x = 0, 1