Question
olve this equation and select the choice that ontains all of the solutions. \[ x^{2}-2 x+2=0 \]
Ask by Cervantes Barber. in the United States
Mar 14,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The solutions are \( x = 1 + i \) and \( x = 1 - i \).
Solution
We are given the quadratic equation:
\[
x^2 - 2x + 2 = 0
\]
**Step 1.** Identify the coefficients from the standard form \( ax^2 + bx + c = 0 \):
\[
a = 1,\quad b = -2,\quad c = 2
\]
**Step 2.** Use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Substitute the values:
\[
x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(2)}}{2(1)} = \frac{2 \pm \sqrt{4 - 8}}{2}
\]
**Step 3.** Simplify under the square root:
\[
\sqrt{4 - 8} = \sqrt{-4} = 2i
\]
**Step 4.** Write the solutions:
\[
x = \frac{2 \pm 2i}{2} = 1 \pm i
\]
Thus, the two solutions to the equation are:
\[
x = 1+i \quad \text{and} \quad x = 1-i
\]
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Simplify this solution The Deep Dive
To solve the equation \( x^{2}-2x+2=0 \), we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = -2 \), and \( c = 2 \). Plugging in these values gives: \[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm \sqrt{-4}}{2} = \frac{2 \pm 2i}{2} = 1 \pm i \] Thus, the solutions are \( x = 1 + i \) and \( x = 1 - i \). The choice that contains all the solutions is: \( \{1 + i, 1 - i\} \). Remember, when you’re dealing with complex numbers, they always come in pairs of conjugates when derived from real coefficients!
