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

Resolve:

x2−13x+3+(x−3)(x+3)=(x−6)(x+6) x^2-13x+3+(x-3)(x+3)=(x-6)(x+6)

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Solve
00:06 Let's use the shortened multiplication formulas
00:37 Calculate 3 squared and 6 squared
00:45 Simplify what we can
00:58 Arrange the equation so that the right side equals 0
01:09 Group terms
01:20 Use the shortened multiplication formulas and find two numbers
01:23 whose sum equals minus 13
01:27 and their product equals 30
01:34 These are the numbers
01:45 Find the two possible solutions that make the equation equal zero
01:59 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Resolve:

x2−13x+3+(x−3)(x+3)=(x−6)(x+6) x^2-13x+3+(x-3)(x+3)=(x-6)(x+6)

2

Step-by-step solution

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 x2−13x+3+(x−3)(x+3) x^2 - 13x + 3 + (x-3)(x+3) . Using the difference of squares formula, expand (x−3)(x+3) (x-3)(x+3) as:

(x−3)(x+3)=x2−32=x2−9 (x-3)(x+3) = x^2 - 3^2 = x^2 - 9

Substituting back, the left side becomes:

x2−13x+3+x2−9 x^2 - 13x + 3 + x^2 - 9

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

(x−6)(x+6)=x2−62=x2−36 (x-6)(x+6) = x^2 - 6^2 = x^2 - 36

Step 2: Simplify both sides.

Combine terms on the left side:

x2−13x+3+x2−9=2x2−13x−6 x^2 - 13x + 3 + x^2 - 9 = 2x^2 - 13x - 6

The right side remains:

x2−36 x^2 - 36

The equation becomes:

2x2−13x−6=x2−36 2x^2 - 13x - 6 = x^2 - 36

Subtract x2 x^2 from both sides to simplify further:

2x2−13x−6−x2=−36 2x^2 - 13x - 6 - x^2 = -36 x2−13x−6=−36 x^2 - 13x - 6 = -36

Step 3: Solve for x x .

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

x2−13x−6+36=0 x^2 - 13x - 6 + 36 = 0 x2−13x+30=0 x^2 - 13x + 30 = 0

Factor the quadratic:

(x−3)(x−10)=0 (x - 3)(x - 10) = 0

Setting each factor to zero gives:

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

These lead to the solutions:

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

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

3

Final Answer

10,3

Practice Quiz

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Solve:

\( (2+x)(2-x)=0 \)

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