(x%2B2)%2B4(x-2)(x%2B1)%2B2(x-2)(x%2B2)%3D0.jpeg "(8-5x)(x+2)+4(x-2)(x+1)+2(x-2)(x+2)=0")
Solving the Quadratic Equation: (8-5x)(x+2)+4(x-2)(x+1)+2(x-2)(x+2)=0
This article will guide you through the steps of solving the given quadratic equation.
1. Expand the Equation:
First, we need to expand the equation by multiplying out the brackets:
- (8-5x)(x+2) = 8x + 16 - 5x² - 10x = -5x² - 2x + 16
- 4(x-2)(x+1) = 4(x² - x - 2) = 4x² - 4x - 8
- 2(x-2)(x+2) = 2(x² - 4) = 2x² - 8
Now we can substitute these back into the original equation:
-5x² - 2x + 16 + 4x² - 4x - 8 + 2x² - 8 = 0
2. Simplify the Equation:
Combine the like terms to simplify the equation:
x² - 6x = 0
3. Factor the Equation:
Factor out the common factor x:
x(x - 6) = 0
4. Solve for x:
For the product of two factors to equal zero, at least one of them must be zero. Therefore, we have two possible solutions:
- x = 0
- x - 6 = 0 => x = 6
5. Solution:
The solutions to the quadratic equation (8-5x)(x+2)+4(x-2)(x+1)+2(x-2)(x+2)=0 are x = 0 and x = 6.