Question

Determine the zeros of the quadratic function y = x²+162 - 39. (When listing the smaller zero, remember that negative values are always smaller than positive values.) (1 point) smaller zero: greater zero: =

Answer 1

To find the zeros of the quadratic function \( y = x^2 + 162 - 39 \), we first simplify the equation:

\[
y = x^2 + 123
\]

Next, to find the zeros, we set \( y = 0 \):

\[
x^2 + 123 = 0
\]

Now, we can isolate \( x^2 \):

\[
x^2 = -123
\]

Taking the square root of both sides gives:

\[
x = \pm \sqrt{-123}
\]

Because the square root of a negative number involves imaginary numbers, we can express it as:

\[
x = \pm \sqrt{123}i
\]

Now, we can simplify \( \sqrt{123} \):

\[
\sqrt{123} = \sqrt{3 \times 41} = \sqrt{3} \cdot \sqrt{41}
\]

Thus, we can express the zeros as:

\[
x = \pm \sqrt{123}i
\]

We can now denote the two zeros:

- The smaller zero (the one with the negative sign) is:
\[
-\sqrt{123}i
\]

- The greater zero (the one with the positive sign) is:
\[
\sqrt{123}i
\]

So the final answers are:
- smaller zero: \( -\sqrt{123}i \)
- greater zero: \( \sqrt{123}i \)