(x^2+6x-7)(2x^2-5x-3)=0

Jasonbradley

2 min read

·

Jun 17, 2024

23

Share

Solving the Equation (x^2 + 6x - 7)(2x^2 - 5x - 3) = 0

This equation represents a polynomial equation where the product of two quadratic expressions equals zero. To solve it, we can use the Zero Product Property:

If the product of two or more factors is zero, then at least one of the factors must be zero.

Applying this property to our equation, we get:

x² + 6x - 7 = 0 OR 2x² - 5x - 3 = 0

Now we have two separate quadratic equations to solve.

Solving x² + 6x - 7 = 0

We can solve this equation by factoring:

1. Find two numbers that add up to 6 and multiply to -7. These numbers are 7 and -1.
2. Rewrite the equation: (x + 7)(x - 1) = 0
3. Apply the Zero Product Property:
   • x + 7 = 0 => x = -7
   • x - 1 = 0 => x = 1

Therefore, the solutions for this quadratic equation are x = -7 and x = 1.

Solving 2x² - 5x - 3 = 0

We can solve this equation using the quadratic formula:

x = (-b ± √(b² - 4ac)) / 2a

Where:

• a = 2
• b = -5
• c = -3

Substituting the values into the formula:

x = (5 ± √((-5)² - 4 * 2 * -3)) / (2 * 2) x = (5 ± √(49)) / 4 x = (5 ± 7) / 4

Therefore, the solutions for this quadratic equation are:

• x = 3
• x = -1/2

Final Solution

Combining the solutions from both quadratic equations, the complete solution set for the original equation (x² + 6x - 7)(2x² - 5x - 3) = 0 is:

x = -7, x = 1, x = 3, x = -1/2