Math Problem Statement

Solve #5 – 6 by the Square Root Property 5. (x + 8)2 = 52 6. (2x – 3)2 + 4 = 29

Solution

Let's solve the equations using the Square Root Property.

Problem 5: (x+8)2=52(x + 8)^2 = 52

  1. Start by taking the square root of both sides: (x+8)2=±52\sqrt{(x + 8)^2} = \pm \sqrt{52} Simplifying, we get: x+8=±52x + 8 = \pm \sqrt{52}

  2. Simplify 52\sqrt{52}: 52=4×13=213\sqrt{52} = \sqrt{4 \times 13} = 2\sqrt{13} So the equation becomes: x+8=±213x + 8 = \pm 2\sqrt{13}

  3. Isolate xx by subtracting 8 from both sides: x=8±213x = -8 \pm 2\sqrt{13}

So the solutions are: x=8+213orx=8213x = -8 + 2\sqrt{13} \quad \text{or} \quad x = -8 - 2\sqrt{13}

Problem 6: (2x3)2+4=29(2x - 3)^2 + 4 = 29

  1. Start by isolating the squared term: (2x3)2=294(2x - 3)^2 = 29 - 4 Simplifying, we get: (2x3)2=25(2x - 3)^2 = 25

  2. Take the square root of both sides: (2x3)2=±25\sqrt{(2x - 3)^2} = \pm \sqrt{25} Simplifying, we get: 2x3=±52x - 3 = \pm 5

  3. Solve for xx by considering the two possible cases:

    • Case 1: 2x3=52x - 3 = 5 2x=8x=42x = 8 \quad \Rightarrow \quad x = 4
    • Case 2: 2x3=52x - 3 = -5 2x=2x=12x = -2 \quad \Rightarrow \quad x = -1

So the solutions are: x=4orx=1x = 4 \quad \text{or} \quad x = -1

Would you like more details on any of these steps, or do you have any questions?

Related Questions:

  1. What is the Square Root Property and how does it work?
  2. Can you solve (x7)2=81(x - 7)^2 = 81 using the Square Root Property?
  3. How can you simplify square roots like 52\sqrt{52} step-by-step?
  4. What are some common mistakes to avoid when using the Square Root Property?
  5. Can you apply the Square Root Property to solve (3x+2)2=36(3x + 2)^2 = 36?

Tip: When using the Square Root Property, always remember to consider both the positive and negative roots.

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Math Problem Analysis

Mathematical Concepts

Quadratic Equations
Square Root Property

Formulas

Quadratic equation solution using Square Root Property

Theorems

Square Root Property

Suitable Grade Level

High School