x2+22x +c find the value of c that makes the trinomial a perfect square
c = 121
since (x+a)^2 = x2+2a+a^2, you want 2a = 22.
So, plug in a=11 and you have c = a^2 = 121
To find the value of c that makes the trinomial a perfect square, we need to complete the square for the quadratic expression x^2 + 22x + c.
Step 1: Take half of the coefficient of x, square it, and add it to both sides of the equation.
22/2 = 11
11^2 = 121
x^2 + 22x + 121 + c = 121
Step 2: Simplify the equation.
x^2 + 22x + 121 + c = 121
x^2 + 22x + (121 + c) = 121
Step 3: Rearrange the equation.
x^2 + 22x + 121 + c - 121 = 0
x^2 + 22x + c = 0
Now, since we want the expression to be a perfect square, we need the constant term to be the square of half the coefficient of x, which is 11. Therefore, c should be equal to 11^2, which is 121.
Thus, the value of c that makes the trinomial a perfect square is 121.
To find the value of c that makes the trinomial a perfect square, we need to determine the value that completes the square.
To complete the square, we take half of the coefficient of x and square it. In this case, the coefficient of x is 22, so we take half of it, which is 11, and square it, giving us 121. Therefore, the value of c that makes the trinomial a perfect square is 121.
So, c = 121.