The set of values of x for which the inequalities ` x^2 -2x+3 gt 0, 2x^2+4x +3gt 0` hold simultancously , is

Options

A. (1, 2)
B. (2, 5)
C. `(0, infty)`
D. `(-infty, infty)`

Answer

To solve the inequalities \( x^2 - 2x + 3 > 0 \) and \( 2x^2 + 4x + 3 > 0 \) simultaneously, we will analyze each quadratic expression step by step.

### Step 1: Analyze the first inequality \( x^2 - 2x + 3 > 0 \)

1. Identify the coefficients:
- \( A = 1 \)
- \( B = -2 \)
- \( C = 3 \)

2. Calculate the discriminant \( D \):
\[ D = B^2 - 4AC = (-2)^2 - 4 \cdot 1 \cdot 3 = 4 - 12 = -8 \]

3. Since \( D < 0 \), the quadratic does not have real roots and opens upwards (as \( A > 0 \)). Therefore, \( x^2 - 2x + 3 > 0 \) for all real values of \( x \).

### Step 2: Analyze the second inequality \( 2x^2 + 4x + 3 > 0 \)

1. Identify the coefficients:
- \( A = 2 \)
- \( B = 4 \)
- \( C = 3 \)

2. Calculate the discriminant \( D \):
\[ D = B^2 - 4AC = 4^2 - 4 \cdot 2 \cdot 3 = 16 - 24 = -8 \]

3. Since \( D < 0 \), this quadratic also does not have real roots and opens upwards (as \( A > 0 \)). Therefore, \( 2x^2 + 4x + 3 > 0 \) for all real values of \( x \).

### Step 3: Combine the results

Both inequalities hold true for all real values of \( x \). Thus, the set of values of \( x \) for which both inequalities hold simultaneously is:

\[ \text{All real numbers } x \in \mathbb{R} \]

### Final Answer:
The set of values of \( x \) for which the inequalities \( x^2 - 2x + 3 > 0 \) and \( 2x^2 + 4x + 3 > 0 \) hold simultaneously is \( \mathbb{R} \).

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