High School

Find the zeros of the function [tex]f(x) = 1.1x^2 + 13x + 35.8[/tex]. Round values to the nearest thousandth, if necessary.

Answer :

To find the zeros of the function [tex]\( f(x) = 1.1x^2 + 13x + 35.8 \)[/tex], you can use the quadratic formula, which is:

[tex]\[
x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}
\][/tex]

Here, the coefficients are:
- [tex]\( a = 1.1 \)[/tex]
- [tex]\( b = 13 \)[/tex]
- [tex]\( c = 35.8 \)[/tex]

Let's break it down step by step:

1. Calculate the Discriminant:
The discriminant is given by [tex]\( b^2 - 4ac \)[/tex].

[tex]\[
b^2 = 13^2 = 169
\][/tex]

[tex]\[
4ac = 4 \times 1.1 \times 35.8 = 157.52
\][/tex]

So, the discriminant is:

[tex]\[
169 - 157.52 = 11.48
\][/tex]

2. Calculate the Square Root of the Discriminant:

[tex]\[
\sqrt{11.48} \approx 3.388
\][/tex]

3. Find the Two Possible Solutions:

Using the quadratic formula:

[tex]\[
x_1 = \frac{{-b + \sqrt{{b^2 - 4ac}}}}{2a} = \frac{{-13 + 3.388}}{2 \times 1.1}
\][/tex]

[tex]\[
x_1 = \frac{{-13 + 3.388}}{2.2} \approx \frac{{-9.612}}{2.2} \approx -4.369
\][/tex]

[tex]\[
x_2 = \frac{{-b - \sqrt{{b^2 - 4ac}}}}{2a} = \frac{{-13 - 3.388}}{2 \times 1.1}
\][/tex]

[tex]\[
x_2 = \frac{{-13 - 3.388}}{2.2} \approx \frac{{-16.388}}{2.2} \approx -7.449
\][/tex]

4. Round the Solutions:

- [tex]\( x_1 \approx -4.369 \)[/tex]
- [tex]\( x_2 \approx -7.449 \)[/tex]

Therefore, the zeros of the function are approximately [tex]\(-4.369\)[/tex] and [tex]\(-7.449\)[/tex].