Certainly! Let's go through the questions step-by-step to address the mistakes, corrections, and the solution for a new quadratic equation.
### 1. Describe the mistake each student made.
Student 1:
- Mistake: In Step 5, the student mistakenly introduces an imaginary component (i) in the solution [tex]$x = \frac{6 - 2i}{2}$[/tex].
- The correct operation should involve only real numbers because the discriminant (the value under the square root sign) in Step 3 results in 4, which is a perfect square and positive. Hence, the solutions should not involve imaginary numbers.
Student 2:
- Mistake: In Step 6, one of the solutions was incorrectly calculated as imaginary: [tex]$x = 3 - i$[/tex].
- Both solutions should be real numbers, considering that [tex]$\sqrt{4}$[/tex] results in 2, which is a real number.
### 2. Explain what each student needs to do to fix their mistake.
Student 1:
- Correction: In step 5, the student should perform the correct arithmetic after taking the square root of 4.
- The step should be [tex]$x = \frac{6+2}{2}$[/tex] and [tex]$x = \frac{6-2}{2}$[/tex].
Student 2:
- Correction: In step 6, correct the solutions by properly simplifying the fractions after solving the square root value.
- The correct simplified solutions should be [tex]$x = 4$[/tex] and [tex]$x = 3$[/tex].
### 3. Create your own quadratic equation, and explain how to use the quadratic formula to solve it.
Quadratic Equation:
Let's use the quadratic equation [tex]\(x^2 - 5x + 6 = 0\)[/tex].
Solution Using the Quadratic Formula:
The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
In our equation, [tex]\(a = 1\)[/tex], [tex]\(b = -5\)[/tex], and [tex]\(c = 6\)[/tex].
Steps:
1. Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = (-5)^2 - 4 \cdot 1 \cdot 6 \][/tex]
[tex]\[ \Delta = 25 - 24 \][/tex]
[tex]\[ \Delta = 1 \][/tex]
2. Calculate the square root of the discriminant:
[tex]\[ \sqrt{\Delta} = \sqrt{1} = 1 \][/tex]
3. Plug the values into the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
[tex]\[ x = \frac{-(-5) \pm 1}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{5 \pm 1}{2} \][/tex]
4. Calculate the two possible solutions:
[tex]\[ x_1 = \frac{5 + 1}{2} = \frac{6}{2} = 3 \][/tex]
[tex]\[ x_2 = \frac{5 - 1}{2} = \frac{4}{2} = 2 \][/tex]
Solutions:
The solutions to the quadratic equation [tex]\(x^2 - 5x + 6 = 0\)[/tex] are [tex]\(x = 3\)[/tex] and [tex]\(x = 2\)[/tex].
