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solve for $j$. $j^{2}-23j + 22 = 0$ write each solution as an integer, …

Question

solve for $j$.
$j^{2}-23j + 22 = 0$
write each solution as an integer, proper fraction, or improper fraction in simplest form. if there are multiple solutions, separate them with commas.
$j = \square$
save answer

Explanation:

Step1: Factor the quadratic equation

Find two integers that multiply to 22 and add to -23. The integers are -1 and -22, so:
$j^2 - 23j + 22 = (j - 1)(j - 22) = 0$

Step2: Set each factor equal to 0

For the first factor:
$j - 1 = 0$
For the second factor:
$j - 22 = 0$

Step3: Solve for j in each equation

Solve $j - 1 = 0$:
$j = 1$
Solve $j - 22 = 0$:
$j = 22$

Answer:

$1, 22$