Questions: Use the zero product property to find the solutions to the equation (x+2)(x+3)=12 x=-1 or x=6 x=-6 or x=1 x=-3 or x=-2 x=2 or x=3

Transcript text: ing Quadratic Equations by toring - Assignment d: Aug 25 at 11:48pm iz Instructions Question 5 0.5 pts Use the zero product property to find the solutions to the equation $(x+2)(x+3)=12$ $x=-1$ or $x=6$ $x=-6$ or $x=1$ $x=-3$ or $x=-2$ $x=2$ or $x=3$ Next Previous

Solution

Solution Steps

To solve the equation $(x+2)(x+3)=12$ using the zero product property, we first need to set the equation to zero by moving 12 to the left side. Then, we expand and simplify the equation to form a standard quadratic equation. Finally, we solve for $x$ using the quadratic formula or factoring.

Step 1: Set the Equation to Zero

Given the equation $(x+2)(x+3)=12$, we first set it to zero by moving 12 to the left side: \[ (x+2)(x+3) - 12 = 0 \]

Step 2: Expand and Simplify

Next, we expand the left-hand side: \[ (x+2)(x+3) - 12 = x^2 + 5x + 6 - 12 = x^2 + 5x - 6 \] So, the equation becomes: \[ x^2 + 5x - 6 = 0 \]

Step 3: Solve the Quadratic Equation

We solve the quadratic equation $x^2 + 5x - 6 = 0$ using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where $a = 1$, $b = 5$, and $c = -6$.

Step 4: Calculate the Discriminant

Calculate the discriminant: \[ \Delta = b^2 - 4ac = 5^2 - 4 \cdot 1 \cdot (-6) = 25 + 24 = 49 \]

Step 5: Find the Roots

Using the quadratic formula: \[ x = \frac{-5 \pm \sqrt{49}}{2 \cdot 1} = \frac{-5 \pm 7}{2} \] Thus, the solutions are: \[ x = \frac{-5 + 7}{2} = 1 \quad \text{and} \quad x = \frac{-5 - 7}{2} = -6 \]