What are the solutions to [tex][tex]$2x^2 + 7x + 2 \geq 0$[/tex][/tex]?

A. [tex]\frac{-7 - \sqrt{33}}{4} \ \textless \ x \ \textless \ \frac{-7 + \sqrt{33}}{4}[/tex]

B. [tex]x \ \textless \ \frac{-7 - \sqrt{33}}{4}[/tex] or [tex]x \ \textgreater \ \frac{-7 + \sqrt{33}}{4}[/tex]

C. [tex]x \leq \frac{-7 - \sqrt{33}}{4}[/tex] or [tex]x \geq \frac{-7 + \sqrt{33}}{4}[/tex]

D. [tex]\frac{-7 - \sqrt{33}}{4} \leq x \leq \frac{-7 + \sqrt{33}}{4}[/tex]



Answer :

To solve the inequality [tex]\(2x^2 + 7x + 2 \geq 0\)[/tex], we need to follow these steps:

1. Identify the quadratic function and find its discriminant:
The given quadratic inequality is [tex]\(2x^2 + 7x + 2 \geq 0\)[/tex].
The standard form of a quadratic equation is [tex]\(ax^2 + bx + c\)[/tex], where [tex]\(a = 2\)[/tex], [tex]\(b = 7\)[/tex], and [tex]\(c = 2\)[/tex].

2. Calculate the discriminant:
The discriminant [tex]\(\Delta\)[/tex] of the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] is given by [tex]\(\Delta = b^2 - 4ac\)[/tex].
So, [tex]\(\Delta = 7^2 - 4 \cdot 2 \cdot 2 = 49 - 16 = 33\)[/tex].

3. Find the roots of the quadratic equation:
The roots of the quadratic equation are given by:
[tex]\[ x_{1,2} = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substituting the values, we get:
[tex]\[ x_1 = \frac{-7 - \sqrt{33}}{4} \][/tex]
[tex]\[ x_2 = \frac{-7 + \sqrt{33}}{4} \][/tex]

4. Determine the intervals where the inequality holds:
Since the coefficient of [tex]\(x^2\)[/tex] (which is 2) is positive, the parabola opens upwards. Therefore, the quadratic expression [tex]\(2x^2 + 7x + 2\)[/tex] will be greater than or equal to zero outside the interval [tex]\([x_1, x_2]\)[/tex].

5. Write down the solution to the inequality:
Based on the interval analysis, the quadratic inequality [tex]\(2x^2 + 7x + 2 \geq 0\)[/tex] is satisfied for [tex]\(x \leq \frac{-7 - \sqrt{33}}{4}\)[/tex] or [tex]\(x \geq \frac{-7 + \sqrt{33}}{4}\)[/tex].

Among the given choices, this corresponds to option C:

[tex]\[ \boxed{x \leq \frac{-7 - \sqrt{33}}{4} \text{ or } x \geq \frac{-7 + \sqrt{33}}{4}} \][/tex]

Other Questions