To solve the quadratic equation \( j^2 - 18j + 17 = 0 \), we can use the quadratic formula:
\[ j = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For the given equation, we have:
- \( a = 1 \)
- \( b = -18 \)
- \( c = 17 \)
Now we can substitute these values into the formula:
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Calculate the discriminant (\( b^2 - 4ac \)): \[ b^2 = (-18)^2 = 324 \] \[ 4ac = 4 \cdot 1 \cdot 17 = 68 \] \[ b^2 - 4ac = 324 - 68 = 256 \]
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Now we can find \( j \) using the quadratic formula: \[ j = \frac{-(-18) \pm \sqrt{256}}{2 \cdot 1} \] \[ j = \frac{18 \pm 16}{2} \]
Now we compute the two potential solutions:
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First solution: \[ j = \frac{18 + 16}{2} = \frac{34}{2} = 17 \]
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Second solution: \[ j = \frac{18 - 16}{2} = \frac{2}{2} = 1 \]
Thus, the solutions for \( j \) are:
\[ j = 17, 1 \]
So the final answer is:
\[ \boxed{17, 1} \]