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