Solve for j j^2-17j+72=0

$j^{2} - 17 j + 72 = 0$

Factor $j^{2} - 17 j + 72$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $72$ and whose sum is $- 17$ .

$- 9, - 8$

Write the factored form using these integers.

$(j - 9) (j - 8) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$j - 9 = 0$

$j - 8 = 0$

Set $j - 9$ equal to $0$ and solve for $j$ .

Set $j - 9$ equal to $0$ .

$j - 9 = 0$

Add $9$ to both sides of the equation.

$j = 9$

Set $j - 8$ equal to $0$ and solve for $j$ .

Set $j - 8$ equal to $0$ .

$j - 8 = 0$

Add $8$ to both sides of the equation.

$j = 8$

The final solution is all the values that make $(j - 9) (j - 8) = 0$ true.

$j = 9, 8$

Solve for j j^2-17j+72=0

Related Questions

Support LCMGCF.com by disabling your adblocker.