(x-3)(x+5)=20

2 min read Jun 17, 2024
(x-3)(x+5)=20

Solving the Equation (x-3)(x+5) = 20

This equation involves a product of two binomials, and we need to solve for the value(s) of x that satisfy the equation. Here's how we can approach it:

1. Expand the equation:

First, we need to multiply out the left-hand side of the equation. Using the distributive property (or FOIL method), we get:

(x-3)(x+5) = x² + 2x - 15

Now our equation becomes:

x² + 2x - 15 = 20

2. Move all terms to one side:

To solve a quadratic equation, we need to have it in the standard form: ax² + bx + c = 0. Let's rearrange the equation:

x² + 2x - 35 = 0 

3. Factor the quadratic expression:

Now we need to factor the quadratic expression on the left-hand side. We need to find two numbers that multiply to -35 and add up to 2. These numbers are 7 and -5.

Therefore, we can factor the equation as:

(x + 7)(x - 5) = 0

4. Solve for x:

For the product of two factors to be zero, at least one of them must be zero. So, we have two possible solutions:

  • x + 7 = 0 => x = -7
  • x - 5 = 0 => x = 5

Conclusion

Therefore, the solutions to the equation (x-3)(x+5) = 20 are x = -7 and x = 5.

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