Math Problem Statement

Solve the quadratic equations given in the image, including equations like (x - 4)^2 = 20, (3x + 2 / 5)^2 = 144, and more.

Solution

Let's solve the equations visible in the image step by step:


6. (x4)2=20(x - 4)^2 = 20

Solution:

  1. Take the square root of both sides: x4=±20=±25x - 4 = \pm \sqrt{20} = \pm 2\sqrt{5}
  2. Solve for xx: x=4±25x = 4 \pm 2\sqrt{5} Thus, the solutions are: x=4+25,x=425x = 4 + 2\sqrt{5}, \quad x = 4 - 2\sqrt{5}

7. (3x+25)2=144\left(\frac{3x + 2}{5}\right)^2 = 144

Solution:

  1. Take the square root of both sides: 3x+25=±144=±12\frac{3x + 2}{5} = \pm \sqrt{144} = \pm 12
  2. Multiply both sides by 5: 3x+2=±603x + 2 = \pm 60
  3. Solve for xx by considering both cases:
    • For 3x+2=603x + 2 = 60: 3x=58    x=5833x = 58 \implies x = \frac{58}{3}
    • For 3x+2=603x + 2 = -60: 3x=62    x=6233x = -62 \implies x = \frac{-62}{3} Thus, the solutions are: x=583,x=623x = \frac{58}{3}, \quad x = \frac{-62}{3}

8. (2x6)2+7=71(2x - 6)^2 + 7 = 71

Solution:

  1. Subtract 7 from both sides: (2x6)2=64(2x - 6)^2 = 64
  2. Take the square root of both sides: 2x6=±82x - 6 = \pm 8
  3. Solve for xx in both cases:
    • For 2x6=82x - 6 = 8: 2x=14    x=72x = 14 \implies x = 7
    • For 2x6=82x - 6 = -8: 2x=2    x=12x = -2 \implies x = -1 Thus, the solutions are: x=7,x=1x = 7, \quad x = -1

9. (5x+13)22=1\left(\frac{5x + 1}{3}\right)^2 - 2 = 1

Solution:

  1. Add 2 to both sides: (5x+13)2=3\left(\frac{5x + 1}{3}\right)^2 = 3
  2. Take the square root of both sides: 5x+13=±3\frac{5x + 1}{3} = \pm \sqrt{3}
  3. Multiply both sides by 3: 5x+1=±335x + 1 = \pm 3\sqrt{3}
  4. Solve for xx: 5x=1±335x = -1 \pm 3\sqrt{3} x=1±335x = \frac{-1 \pm 3\sqrt{3}}{5} Thus, the solutions are: x=1+335,x=1335x = \frac{-1 + 3\sqrt{3}}{5}, \quad x = \frac{-1 - 3\sqrt{3}}{5}

10. 9(2x+4)2106=3\frac{9(-2x + 4)^2}{10} - 6 = 3

Solution:

  1. Add 6 to both sides: 9(2x+4)210=9\frac{9(-2x + 4)^2}{10} = 9
  2. Multiply both sides by 10: 9(2x+4)2=909(-2x + 4)^2 = 90
  3. Divide by 9: (2x+4)2=10(-2x + 4)^2 = 10
  4. Take the square root of both sides: 2x+4=±10-2x + 4 = \pm \sqrt{10}
  5. Solve for xx: 2x=4±10    x=2102-2x = -4 \pm \sqrt{10} \implies x = 2 \mp \frac{\sqrt{10}}{2} Thus, the solutions are: x=2+102,x=2102x = 2 + \frac{\sqrt{10}}{2}, \quad x = 2 - \frac{\sqrt{10}}{2}

Do you need further details or explanations?
Here are 5 related questions you might be interested in:

  1. How do you solve quadratic equations using the quadratic formula?
  2. What are the rules for dealing with square roots when solving equations?
  3. How can we check the validity of solutions for quadratic equations?
  4. What are the steps for solving equations with fractions or radicals?
  5. What is the method of completing the square for solving quadratic equations?

Tip: Always check if the solutions make sense by substituting them back into the original equation!

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Math Problem Analysis

Mathematical Concepts

Quadratic Equations
Algebra
Square Root Method
Linear Equations

Formulas

(x - h)^2 = k -> x = h ± sqrt(k)
ax^2 + bx + c = 0
Linear solution: ax + b = 0 -> x = -b/a

Theorems

Properties of Squares
Square Root Theorem

Suitable Grade Level

Grades 8-11