Question

What are the zeros for the function f(x) = 5x2 + 2x + 1 and how many times does the graph cross the x-axis?

Answer 1

For a quadratic function

`ax^2+bx+c=0`

The x-coordinates for the roots can be found using the Bhaskara formula:

`x=\frac{-b\pm\sqrt[]{b^2-4\cdot a\cdot c}}{2a}`

So, to solve this question, follow the steps below.

Step 01: Find a, b and c.

For the equation

`f(x)=5x^2+2x+1`

a = 5

b = 2

c = 1

Step 02: Substitute the values in the Bhaskara formula to find x.

`\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}`

Since i² = -1, you can substitute -16 by 16*i²:

`\begin{gathered} x=\frac{-2\pm\sqrt[]{16\cdot i^2}}{10}=\frac{-2\pm\sqrt[]{16}\cdot\sqrt[]{i^2}}{10} \\ x=(-2\pm4\cdot i)/(10) \\ x=(-1\pm2i)/(5) \end{gathered}`

The roots are:

`\begin{gathered} x_1=(-1+2i)/(5) \\ \text{and} \\ x_2=(-1-2i)/(5) \end{gathered}`

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:

`x=(-1\pm2i)/(5),\text{ never}`