Given the polynomial:
[tex]y=a^2+16a+64[/tex]
The x-intercepts are the points in which the polynomial crosses the x-axis. That is, are the points (x, 0).
To find these points, use the quadratic formula according to the steps below.
Step 01: Substitute the values in the quadratic formula.
For a equation y = ax² + bx + c, the x-intercepts are:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
In this question,
a = 1
b = 16
c = 64
Substituting the values:
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-16\pm\sqrt[]{16^2-4\cdot1\cdot64}}{2\cdot1} \end{gathered}[/tex]
Step 02: Solve the equation above.
Begin by solving the square and the multiplications:
[tex]x=\frac{-16\pm\sqrt[]{256^{}-256}}{2}[/tex]
Now, subtract the values inside the root.
[tex]x=\frac{-16\pm\sqrt[]{0}}{2}[/tex]
Solve the root:
[tex]x=\frac{-16\pm0}{2}[/tex]
And find the values of x:
[tex]\begin{gathered} x_1=\frac{-16-0}{2}=-\frac{16}{2}=-8 \\ x_2=\frac{-16+0}{2}=-\frac{16}{2}=-8 \end{gathered}[/tex]
Answer: The x-intercepts are: (-8, 0), (-8, 0).
