Math Problem Statement

(x + 5)^2 - 3(x + 5) - 4 = 0

Solution

To solve the equation (x+5)23(x+5)4=0(x + 5)^2 - 3(x + 5) - 4 = 0, let's proceed with a substitution method.

Step 1: Substitute

Let y=x+5y = x + 5. This substitution simplifies the equation. Rewriting the equation in terms of yy gives us:

y23y4=0y^2 - 3y - 4 = 0

Step 2: Factor the Quadratic Equation

We now have a quadratic equation in yy:

y23y4=0y^2 - 3y - 4 = 0

This equation can be factored as:

(y4)(y+1)=0(y - 4)(y + 1) = 0

Step 3: Solve for yy

Setting each factor to zero, we get:

  1. y4=0y=4y - 4 = 0 \Rightarrow y = 4
  2. y+1=0y=1y + 1 = 0 \Rightarrow y = -1

Step 4: Substitute Back for xx

Recall that y=x+5y = x + 5, so we substitute back to find xx:

  1. If y=4y = 4: x+5=4x=1x + 5 = 4 \Rightarrow x = -1

  2. If y=1y = -1: x+5=1x=6x + 5 = -1 \Rightarrow x = -6

Final Solution

The solutions to the equation are:

x=1andx=6x = -1 \quad \text{and} \quad x = -6

Would you like further details or have any questions?

Here are some related questions you might explore:

  1. How do you determine when substitution is useful in solving equations?
  2. Can you apply this substitution method to other types of equations?
  3. What are other methods to solve quadratic equations besides factoring?
  4. How does completing the square work for similar equations?
  5. Can this equation be solved by graphing instead?

Tip: Always check your solutions by substituting them back into the original equation to ensure they satisfy it.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Substitution Method

Formulas

Standard form of a quadratic equation ax^2 + bx + c = 0

Theorems

Zero Product Property

Suitable Grade Level

Grades 9-10