Find the Roots (Zeros) f(t)=t^2-13t+36

$f (t) = t^{2} - 13 t + 36$

Step 1

Set $t^{2} - 13 t + 36$ equal to $0$ .

$t^{2} - 13 t + 36 = 0$

Step 2

Solve for $x$ .

Factor $t^{2} - 13 t + 36$ 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 $36$ and whose sum is $- 13$ .

$- 9, - 4$

Write the factored form using these integers.

$(t - 9) (t - 4) = 0$

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

$t - 9 = 0$

$t - 4 = 0$

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

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

$t - 9 = 0$

Add $9$ to both sides of the equation.

$t = 9$

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

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

$t - 4 = 0$

Add $4$ to both sides of the equation.

$t = 4$

The final solution is all the values that make $(t - 9) (t - 4) = 0$ true.

$t = 9, 4$

Step 3

Find the Roots (Zeros) f(t)=t^2-13t+36

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