For a quadratic function
[tex]ax^2+bx+c=0[/tex]
The x-coordinates for the roots can be found using the Bhaskara formula:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4\cdot a\cdot c}}{2a}[/tex]
So, to solve this question, follow the steps below.
Step 01: Find a, b and c.
For the equation
[tex]f(x)=5x^2+2x+1[/tex]
a = 5
b = 2
c = 1
Step 02: Substitute the values in the Bhaskara formula to find x.
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4\cdot a\cdot c}}{2a} \\ x=\frac{-2\pm\sqrt[]{2^2-4\cdot5\cdot1}}{2\cdot5} \\ x=\frac{-2\pm\sqrt[]{4-20}}{10} \\ x=\frac{-2\pm\sqrt[]{-16}}{10} \end{gathered}[/tex]
Since i² = -1, you can substitute -16 by 16*i²:
[tex]\begin{gathered} x=\frac{-2\pm\sqrt[]{16\cdot i^2}}{10}=\frac{-2\pm\sqrt[]{16}\cdot\sqrt[]{i^2}}{10} \\ x=\frac{-2\pm4\cdot i}{10} \\ x=\frac{-1\pm2i}{5} \end{gathered}[/tex]
The roots are:
[tex]\begin{gathered} x_1=\frac{-1+2i}{5} \\ \text{and} \\ x_2=\frac{-1-2i}{5} \end{gathered}[/tex]
Step 03: Evaluate where the equation crosses the x-axis.
When the value inside the root is negative, it means that the equation does not cresses the x-axis.
Also, you can graph the equation to observe it:
Answer:
[tex]x=\frac{-1\pm2i}{5},\text{ never}[/tex]
