Math Problem Statement

x22x5=0 The solution set is ​{enter your response here​}. ​(Simplify your answer. Type an integer or decimal rounded to two decimal places as needed. Use a comma to separate answers as​ needed.)x^2-\sqrt2x-5=0\text{ The solution set is ​\textbraceleft enter your response here​\textbraceright. ​(Simplify your answer. Type an integer or decimal rounded to two decimal places as needed. Use a comma to separate answers as​ needed.)}

Solution

To solve the quadratic equation x22x5=0x^2 - \sqrt{2}x - 5 = 0 using the quadratic formula, we start with the standard form of the quadratic equation:

ax2+bx+c=0ax^2 + bx + c = 0

Here, a=1a = 1, b=2b = -\sqrt{2}, and c=5c = -5.

The quadratic formula is:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substitute the values of aa, bb, and cc:

x=(2)±(2)24(1)(5)2(1)x = \frac{-(-\sqrt{2}) \pm \sqrt{(-\sqrt{2})^2 - 4(1)(-5)}}{2(1)}

Simplify this step by step:

  1. Calculate the discriminant:

(2)2=2(-\sqrt{2})^2 = 2

4ac=4(1)(5)=204ac = 4(1)(-5) = -20

b24ac=2+20=22b^2 - 4ac = 2 + 20 = 22

  1. Plug the discriminant back into the quadratic formula:

x=2±222x = \frac{\sqrt{2} \pm \sqrt{22}}{2}

  1. Simplify further:

x1=2+222x_1 = \frac{\sqrt{2} + \sqrt{22}}{2}

x2=2222x_2 = \frac{\sqrt{2} - \sqrt{22}}{2}

  1. Calculate the approximate decimal values (rounded to two decimal places):

x11.41+4.692=6.102=3.05x_1 \approx \frac{1.41 + 4.69}{2} = \frac{6.10}{2} = 3.05

x21.414.692=3.282=1.64x_2 \approx \frac{1.41 - 4.69}{2} = \frac{-3.28}{2} = -1.64

Thus, the solution set is:

{3.05,1.64}\{3.05, -1.64\}

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

Here are 5 relative questions:

  1. What is the discriminant and how does it affect the number of real solutions?
  2. How does the quadratic formula apply to equations where the coefficient aa is not 1?
  3. Can the quadratic formula be used for complex solutions? If so, how?
  4. What are some alternative methods to solve quadratic equations besides the quadratic formula?
  5. How would the solution set change if cc were positive instead of negative?

Tip: Always check the discriminant first before proceeding with the quadratic formula, as it will tell you whether real solutions exist.

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Math Problem Analysis

Mathematical Concepts

Quadratic Equations
Quadratic Formula
Discriminant

Formulas

Quadratic formula

Theorems

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Suitable Grade Level

Grades 10-12