`2x^(2) + 3x - alpha - 0 " has roots "-2 and beta " while the equation "x^(2) - 3mx + 2m^(2) = 0 " has both roots positive, where " alpha gt 0 and beta gt 0.`
The equation `|1-x|+ x^(2) = 5` has

Options

A. a rational root and an irrrational root
B. two rational roots
C. two irrational roots
D. no real roots

Explanation

To solve the equation \( |1 - x| + x^2 = 5 \), we will consider two cases based on the definition of the absolute value.

### Step 1: Case 1 - \( x < 1 \)
In this case, \( |1 - x| = 1 - x \). Therefore, the equation becomes:
\[ 1 - x + x^2 = 5 \]
Rearranging this gives:
\[ x^2 - x + 1 - 5 = 0 \implies x^2 - x - 4 = 0 \]

### Step 2: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -1, c = -4 \):
\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 + 16}}{2} = \frac{1 \pm \sqrt{17}}{2} \]
This gives us two potential solutions:
\[ x_1 = \frac{1 + \sqrt{17}}{2}, \quad x_2 = \frac{1 - \sqrt{17}}{2} \]

### Step 3: Check if the roots are valid for \( x < 1 \)
Now, we need to check if these roots satisfy the condition \( x < 1 \):
- For \( x_1 = \frac{1 + \sqrt{17}}{2} \): Since \( \sqrt{17} \approx 4.123 \), we have \( x_1 \approx \frac{5.123}{2} \approx 2.5615 \), which does not satisfy \( x < 1 \).
- For \( x_2 = \frac{1 - \sqrt{17}}{2} \): This will be negative since \( \sqrt{17} > 1 \). Thus, \( x_2 < 1 \) is valid.

### Step 4: Case 2 - \( x \geq 1 \)
In this case, \( |1 - x| = x - 1 \). Therefore, the equation becomes:
\[ x - 1 + x^2 = 5 \]
Rearranging gives:
\[ x^2 + x - 6 = 0 \]

### Step 5: Solve the quadratic equation
Again using the quadratic formula:
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 24}}{2} = \frac{-1 \pm \sqrt{25}}{2} = \frac{-1 \pm 5}{2} \]
This gives us:
\[ x_3 = \frac{4}{2} = 2, \quad x_4 = \frac{-6}{2} = -3 \]

### Step 6: Check if the roots are valid for \( x \geq 1 \)
- For \( x_3 = 2 \): This satisfies \( x \geq 1 \).
- For \( x_4 = -3 \): This does not satisfy \( x \geq 1 \).

### Conclusion
Thus, the valid roots of the equation \( |1 - x| + x^2 = 5 \) are:
1. \( x_2 = \frac{1 - \sqrt{17}}{2} \) (valid for \( x < 1 \))
2. \( x_3 = 2 \) (valid for \( x \geq 1 \))

### Final Answer
The equation has **two roots**: one irrational root \( \frac{1 - \sqrt{17}}{2} \) and one rational root \( 2 \).

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