Solve by Factoring z^2+28=11z

$z^{2} + 28 = 11 z$

Step 1

Subtract $11 z$ from both sides of the equation.

$z^{2} + 28 - 11 z = 0$

Step 2

Factor $z^{2} + 28 - 11 z$ 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 $28$ and whose sum is $- 11$ .

$- 7, - 4$

Write the factored form using these integers.

$(z - 7) (z - 4) = 0$

Step 3

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

$z - 7 = 0$

$z - 4 = 0$

Step 4

Set $z - 7$ equal to $0$ and solve for $z$ .

Set $z - 7$ equal to $0$ .

$z - 7 = 0$

Add $7$ to both sides of the equation.

$z = 7$

Step 5

Set $z - 4$ equal to $0$ and solve for $z$ .

Set $z - 4$ equal to $0$ .

$z - 4 = 0$

Add $4$ to both sides of the equation.

$z = 4$

Step 6

The final solution is all the values that make $(z - 7) (z - 4) = 0$ true.

$z = 7, 4$

Solve by Factoring z^2+28=11z

Related Questions

Support LCMGCF.com by disabling your adblocker.