To determine the number of x-intercepts the quadratic equation [tex]\(2x^2 + 11x + 5 = 0\)[/tex] has, we need to calculate its discriminant. The discriminant of a quadratic equation in the form [tex]\(ax^2 + bx + c = 0\)[/tex] is found using the formula:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Here, the coefficients are:
- [tex]\(a = 2\)[/tex]
- [tex]\(b = 11\)[/tex]
- [tex]\(c = 5\)[/tex]
First, substitute the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] into the formula for the discriminant:
[tex]\[
\Delta = 11^2 - 4 \cdot 2 \cdot 5
\][/tex]
Calculate [tex]\(11^2\)[/tex]:
[tex]\[
11^2 = 121
\][/tex]
Next, calculate [tex]\(4 \cdot 2 \cdot 5\)[/tex]:
[tex]\[
4 \cdot 2 = 8
\][/tex]
[tex]\[
8 \cdot 5 = 40
\][/tex]
Subtract [tex]\(40\)[/tex] from [tex]\(121\)[/tex]:
[tex]\[
\Delta = 121 - 40 = 81
\][/tex]
Now, analyze the discriminant:
- If [tex]\(\Delta > 0\)[/tex], the quadratic equation has two distinct real x-intercepts.
- If [tex]\(\Delta = 0\)[/tex], the quadratic equation has exactly one real x-intercept.
- If [tex]\(\Delta < 0\)[/tex], the quadratic equation has no real x-intercepts.
In this case, the discriminant is [tex]\(\Delta = 81\)[/tex].
Since [tex]\(81 > 0\)[/tex], the quadratic equation [tex]\(2x^2 + 11x + 5 = 0\)[/tex] has two distinct real x-intercepts.
Therefore, the number of x-intercepts is:
[tex]\[
\boxed{2}
\][/tex]
