Math Problem Statement

Solve the quadratic equations by extracting the roots: 1. x^2 = 75 2. x^2 = 80 3. 2(x - 5)^2 = 32 4. 3(4x - 1)^2 - 1 = 11 5. (2x - 3)^2 = 18

Solution

The image contains a list of quadratic equations that need to be solved by extracting the roots. Let's go through each problem one by one.

1. x2=75x^2 = 75

To solve for xx, take the square root of both sides: x=±75x = \pm \sqrt{75} Simplifying 75\sqrt{75}: 75=25×3=53\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} So, the solutions are: x=±53x = \pm 5\sqrt{3}

2. x2=80x^2 = 80

Take the square root of both sides: x=±80x = \pm \sqrt{80} Simplifying 80\sqrt{80}: 80=16×5=45\sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5} So, the solutions are: x=±45x = \pm 4\sqrt{5}

3. 2(x5)2=322(x - 5)^2 = 32

First, divide both sides by 2: (x5)2=16(x - 5)^2 = 16 Take the square root of both sides: x5=±4x - 5 = \pm 4 Solve for xx: x=5±4x = 5 \pm 4 The solutions are: x=9orx=1x = 9 \quad \text{or} \quad x = 1

4. 3(4x1)21=113(4x - 1)^2 - 1 = 11

First, add 1 to both sides: 3(4x1)2=123(4x - 1)^2 = 12 Then, divide both sides by 3: (4x1)2=4(4x - 1)^2 = 4 Take the square root of both sides: 4x1=±24x - 1 = \pm 2 Solve for xx:

  1. If 4x1=24x - 1 = 2: 4x=3x=344x = 3 \quad \Rightarrow \quad x = \frac{3}{4}
  2. If 4x1=24x - 1 = -2: 4x=1x=144x = -1 \quad \Rightarrow \quad x = -\frac{1}{4} The solutions are: x=34orx=14x = \frac{3}{4} \quad \text{or} \quad x = -\frac{1}{4}

5. (2x3)2=18(2x - 3)^2 = 18

Take the square root of both sides: 2x3=±182x - 3 = \pm \sqrt{18} Simplify 18\sqrt{18}: 18=9×2=32\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2} Solve for xx: 2x=3±322x = 3 \pm 3\sqrt{2} x=3±322x = \frac{3 \pm 3\sqrt{2}}{2} The solutions are: x=3+322orx=3322x = \frac{3 + 3\sqrt{2}}{2} \quad \text{or} \quad x = \frac{3 - 3\sqrt{2}}{2}


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Relative Questions:

  1. How can you verify the solutions to these quadratic equations?
  2. What are the real-life applications of solving quadratic equations?
  3. How do you solve a quadratic equation when it's not easily factorable?
  4. What is the significance of the discriminant in quadratic equations?
  5. How can you derive the quadratic formula from a standard quadratic equation?

Tip: When solving quadratic equations, always consider checking the 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
Root Extraction

Formulas

x = ±√c
ax^2 + bx + c = 0
Solving by taking square roots

Theorems

Square Root Theorem
Basic Quadratic Equation Solving

Suitable Grade Level

Grades 9-12