Solving the Equation (x-3)(x-5) = -2
This article will guide you through the steps of solving the equation (x-3)(x-5) = -2.
1. Expanding the Equation
First, we need to expand the left side of the equation by multiplying the factors:
(x-3)(x-5) = x² - 5x - 3x + 15 = x² - 8x + 15
Now, our equation becomes: x² - 8x + 15 = -2
2. Rearranging the Equation
To solve a quadratic equation, we need to set it equal to zero. Let's move the constant term from the right side to the left:
x² - 8x + 15 + 2 = 0
This simplifies to: x² - 8x + 17 = 0
3. Solving the Quadratic Equation
Now we have a standard quadratic equation in the form ax² + bx + c = 0. We can solve this using the quadratic formula:
x = (-b ± √(b² - 4ac)) / 2a
In our equation, a = 1, b = -8, and c = 17. Plugging these values into the quadratic formula, we get:
x = (8 ± √((-8)² - 4 * 1 * 17)) / 2 * 1
Simplifying further:
x = (8 ± √(64 - 68)) / 2
x = (8 ± √(-4)) / 2
x = (8 ± 2i) / 2 (where 'i' is the imaginary unit, √-1)
4. Final Solutions
Finally, we get two solutions for x:
- x = 4 + i
- x = 4 - i
Therefore, the solutions to the equation (x-3)(x-5) = -2 are x = 4 + i and x = 4 - i. These are complex solutions, as the discriminant (b² - 4ac) is negative.