Find all real solutions of the equation [tex]3x^2 + 3x + 8 = 0[/tex].

Does the equation have real solutions? [ ] Yes [ ] No

If your answer is Yes, input the solutions:

[tex]x_1 =[/tex] [ ] and [tex]x_2 =[/tex] [ ] with [tex]x_1 \leq x_2[/tex].



Answer :

To find the real solutions of the quadratic equation [tex]\(3x^2 + 3x + 8 = 0\)[/tex], we can follow these steps:

1. Identify the coefficients:
The standard form of a quadratic equation is [tex]\(ax^2 + bx + c = 0\)[/tex]. Here, we have:
[tex]\[ a = 3, \quad b = 3, \quad c = 8 \][/tex]

2. Calculate the discriminant:
The discriminant [tex]\(\Delta\)[/tex] of a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Plugging in the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ \Delta = 3^2 - 4 \cdot 3 \cdot 8 \][/tex]
Simplify it step by step:
[tex]\[ \Delta = 9 - 96 = -87 \][/tex]

3. Analyze the discriminant:
The nature of the roots of the quadratic equation depends on the discriminant:
- If [tex]\(\Delta > 0\)[/tex], there are two distinct real solutions.
- If [tex]\(\Delta = 0\)[/tex], there is one real solution (a repeated root).
- If [tex]\(\Delta < 0\)[/tex], there are no real solutions; the solutions are complex.

Here, the discriminant [tex]\(\Delta = -87\)[/tex] is less than zero. This indicates that there are no real solutions.

Therefore, the quadratic equation [tex]\(3x^2 + 3x + 8 = 0\)[/tex] does not have any real solutions.