The equation `ax^(2) +bx+ c=0,` where a,b,c are the side of a `DeltaABC,` and the equation `x^(2) +sqrt2x+1=0` have a common root. Find measure for `angle C.`

Answer

To solve the problem, we need to find the measure of angle C in triangle ABC given that the sides of the triangle correspond to the coefficients of the quadratic equation \( ax^2 + bx + c = 0 \) and that this equation has a common root with the equation \( x^2 + \sqrt{2}x + 1 = 0 \).

### Step-by-Step Solution:

1. **Identify the given quadratic equation**:
The given quadratic equation is:
\[ x^2 + \sqrt{2}x + 1 = 0 \]

2. **Calculate the discriminant**:
The discriminant \( D \) of a quadratic equation \( Ax^2 + Bx + C = 0 \) is given by:
\[ D = B^2 - 4AC \]
For our equation, \( A = 1 \), \( B = \sqrt{2} \), and \( C = 1 \):
\[ D = (\sqrt{2})^2 - 4 \cdot 1 \cdot 1 = 2 - 4 = -2 \]
Since the discriminant is negative, the roots are complex.

3. **Roots of the quadratic equation**:
The roots of the equation can be expressed as:
\[ x = \frac{-B \pm \sqrt{D}}{2A} = \frac{-\sqrt{2} \pm i\sqrt{2}}{2} \]
This gives us two complex roots:
\[ x_1 = \frac{-\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}, \quad x_2 = \frac{-\sqrt{2}}{2} - i\frac{\sqrt{2}}{2} \]

4. **Common root condition**:
Since the equation \( ax^2 + bx + c = 0 \) has a common root with the above equation, the coefficients \( a, b, c \) must be proportional to the coefficients of the given quadratic equation. Thus, we can write:
\[ \frac{a}{1} = \frac{b}{\sqrt{2}} = \frac{c}{1} = k \]
This implies:
\[ a = k, \quad b = k\sqrt{2}, \quad c = k \]

5. **Using the triangle property**:
Since \( a, b, c \) are the sides of triangle ABC, we can apply the Pythagorean theorem. We know that:
\[ a^2 + c^2 = b^2 \]
Substituting the values of \( a, b, c \):
\[ k^2 + k^2 = (k\sqrt{2})^2 \]
Simplifying this gives:
\[ 2k^2 = 2k^2 \]
This confirms that the triangle is a right triangle.

6. **Determine angles in the triangle**:
Since \( a = c \), angles opposite to these sides are equal. Let the angles opposite to sides \( a \) and \( c \) be \( \theta \). The angle opposite side \( b \) (the hypotenuse) is \( C = 90^\circ \).

7. **Finding angle C**:
The sum of angles in a triangle is \( 180^\circ \):
\[ \theta + \theta + C = 180^\circ \]
Since \( C = 90^\circ \):
\[ 2\theta + 90^\circ = 180^\circ \]
\[ 2\theta = 90^\circ \implies \theta = 45^\circ \]
Therefore, angle \( C \) is:
\[ C = 90^\circ \]

### Final Answer:
The measure of angle C is \( 90^\circ \).