Question
Solve the inequality. \( 3 x^{2}+6 x+4>0 \) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. There is a solution and it is not all real numbers. The solution is B. The solution is all real numbers. C. There is no solution. C inequality or a compound inequality.)
Ask by Wilson Lyons. in the United States
Mar 20,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The solution is all real numbers.
Solution
Since the quadratic function is given by
\[
f(x) = 3x^2 + 6x + 4,
\]
we follow these steps:
1. Identify the coefficients:
\[
a = 3,\quad b = 6,\quad c = 4.
\]
2. Compute the discriminant:
\[
D = b^2 - 4ac = 6^2 - 4(3)(4) = 36 - 48 = -12.
\]
Since \( D < 0 \), the quadratic equation \( 3x^2 + 6x + 4 = 0 \) has no real roots.
3. Since the leading coefficient \( a = 3 \) is positive, the parabola opens upward. With no real roots, the parabola lies entirely above the \( x \)-axis.
4. Thus, the inequality
\[
3x^2 + 6x + 4 > 0
\]
holds for all real numbers.
The correct choice is B: The solution is all real numbers.
Answered by UpStudy AI and reviewed by a Professional Tutor
Like
error msg
Explain 
Simplify this solution Mind Expander
To solve the inequality \( 3x^{2} + 6x + 4 > 0 \), we start by analyzing the quadratic equation \( 3x^{2} + 6x + 4 = 0 \). The discriminant \( b^2 - 4ac \) is \( 6^2 - 4(3)(4) = 36 - 48 = -12 \). Since the discriminant is negative, the quadratic has no real roots. As the coefficient of \( x^{2} \) (which is 3) is positive, the parabola opens upwards, meaning that the expression \( 3x^{2} + 6x + 4 \) is always greater than zero for all real numbers \( x \). So, the correct answer is: B. The solution is all real numbers. And remember, if you ever see a negative discriminant, it's a hint that the parabola is sitting above the x-axis, making it all positive! Isn't math fascinating?
