To find the value of $ c $ that makes the trinomial $ x^2 - 22x + c $ a perfect square, follow these steps:

1. **Identify the coefficient of $ x $:** The coefficient of $ x $ in the trinomial is $-22$.

2. **Divide the coefficient by 2:**
$$
\frac{-22}{2} = -11
$$

3. **Square the result from step 2:**
$$
(-11)^2 = 121
$$

4. **Set $ c $ to the squared result:**
$$
c = 121
$$

Therefore, the value of $ c $ that makes the trinomial $ x^2 - 22x + c $ a perfect square is $ 121 $.

Question

To find the value of $ c $ that makes the trinomial $ x^2 - 22x + c $ a perfect square, follow these steps:

1. **Identify the coefficient of $ x $:** The coefficient of $ x $ in the trinomial is $-22$.

2. **Divide the coefficient by 2:** 
   $$
   \frac{-22}{2} = -11
   $$

3. **Square the result from step 2:**
   $$
   (-11)^2 = 121
   $$

4. **Set $ c $ to the squared result:** 
   $$
   c = 121
   $$

Therefore, the value of $ c $ that makes the trinomial $ x^2 - 22x + c $ a perfect square is $ 121 $.

Knowee AI · Accepted Answer

To find the value of $ c $ that makes the trinomial $ x^2 - 22x + c $ a perfect square, follow these steps:

1. **Identify the coefficient of $ x $:** The coefficient of $ x $ in the trinomial is $-22$.

2. **Divide the coefficient by 2:** 
   $$
   \frac{-22}{2} = -11
   $$

3. **Square the result from step 2:**
   $$
   (-11)^2 = 121
   $$

4. **Set $ c $ to the squared result:** 
   $$
   c = 121
   $$

Therefore, the value of $ c $ that makes the trinomial $ x^2 - 22x + c $ a perfect square is $ 121 $.

Find the value of c that makes the trinomial a perfect square. x2−22x+c