%2B2x%5E2-3x%2B4)%3D6%2F(x%5E2-3x%2B5).jpeg "1/(x^2-3x+3)+2x^2-3x+4)=6/(x^2-3x+5)")
Solving the Equation: 1/(x^2-3x+3) + 2x^2-3x+4 = 6/(x^2-3x+5)
In this article, we will explore the solution to the equation:
$\frac{1}{x^2-3x+3} + 2x^2-3x+4 = \frac{6}{x^2-3x+5}$
To solve this equation, we will start by simplifying the left-hand side of the equation.
Simplifying the Left-Hand Side
First, let's simplify the fraction on the left-hand side:
$\frac{1}{x^2-3x+3} = \frac{1}{(x-1)(x-3)}$
Next, we will add the polynomial terms:
$2x^2-3x+4$
So, the simplified left-hand side is:
$\frac{1}{(x-1)(x-3)} + 2x^2-3x+4$
Simplifying the Right-Hand Side
Now, let's simplify the right-hand side of the equation:
$\frac{6}{x^2-3x+5} = \frac{6}{(x-1)(x-5)}$
Equating the Two Expressions
Now that we have simplified both sides of the equation, we can equate them:
$\frac{1}{(x-1)(x-3)} + 2x^2-3x+4 = \frac{6}{(x-1)(x-5)}$
Solving for x
To solve for x, we can start by multiplying both sides of the equation by the lowest common multiple of the denominators:
$(x-1)(x-3)(x-5) \left(\frac{1}{(x-1)(x-3)} + 2x^2-3x+4\right) = (x-1)(x-3)(x-5)\left(\frac{6}{(x-1)(x-5)}\right)$
Simplifying and combining like terms, we get:
$x^3 - 7x^2 + 19x - 15 = 0$
This is a cubic equation, and solving it will give us the values of x that satisfy the original equation.
Conclusion
In this article, we have simplified the equation 1/(x^2-3x+3) + 2x^2-3x+4 = 6/(x^2-3x+5) and converted it into a cubic equation. Solving this equation will give us the values of x that satisfy the original equation.
