Solve the Quadratic Equation: x² - 13x + 3 + (x-3)(x+3) = (x-6)(x+6)

To solve this quadratic equation, follow these steps:

• Step 1: Expand both sides completely.
• Step 2: Simplify both sides to form a quadratic equation.
• Step 3: Solve the quadratic equation using appropriate methods.

Now, let's work through the steps:

Step 1: Expand the expressions.

The left side is $x^2 - 13x + 3 + (x-3)(x+3)$. Using the difference of squares formula, expand $(x-3)(x+3)$ as:

$(x-3)(x+3) = x^2 - 3^2 = x^2 - 9$

Substituting back, the left side becomes:

$x^2 - 13x + 3 + x^2 - 9$

The right side is $(x-6)(x+6)$. Expand using the difference of squares:

$(x-6)(x+6) = x^2 - 6^2 = x^2 - 36$

Step 2: Simplify both sides.

Combine terms on the left side:

$x^2 - 13x + 3 + x^2 - 9 = 2x^2 - 13x - 6$

The right side remains:

$x^2 - 36$

The equation becomes:

$2x^2 - 13x - 6 = x^2 - 36$

Subtract $x^2$ from both sides to simplify further:

$2x^2 - 13x - 6 - x^2 = -36$ $x^2 - 13x - 6 = -36$

Step 3: Solve for $x$.

Move -36 to the other side to form a standard quadratic equation:

$x^2 - 13x - 6 + 36 = 0$ $x^2 - 13x + 30 = 0$

Factor the quadratic:

$(x - 3)(x - 10) = 0$

Setting each factor to zero gives:

$x - 3 = 0 \quad \text{or} \quad x - 10 = 0$

These lead to the solutions:

$x = 3 \quad \text{and} \quad x = 10$

Thus, the solutions to the equation are $x = 3$ and $x = 10$.