How do you find the c that makes the trinomial #x^2+22x+c# a perfect square?

Question

Ethan Adkins · Accepted Answer

#c = 121#

Which gives #x^2 +22x+121 =(x+11)^2# This is a process called 'Completing the Square' and does exactly what the name implies... To complete means to add what is missing You are trying to create a perfect square, in this case the square of a binomial. In #1x^2 + color(red)(b)x + c," "# if this is a perfect square there is always a specific relationship between #b and c#.... 'Half of #color(red)(b)#, squared, will give the value of #c#' This is #c= (color(red)(b)/2)^2# In #1x^2+ color(red)(22)x + ???" "rarr ??? = (color(red)(22)/2)^2 = 11^2 =121# The trinomial will therefore be #x^2 +22x+121#, which factorises as #(x+11)^2# Note that to do this, the coefficient of #x^2# must be #1#

Nathan Axel · Answer

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

1. Identify the coefficient of the linear term, which is $22$.
2. Divide the coefficient of the linear term by $2$, then square the result. This gives us $\left(\frac{22}{2}\right)^2 = 121$.
3. The constant $c$ that makes the trinomial a perfect square is equal to the squared result from step 2, so $c = 121$.

Abigail Allen · Answer

undefined To find the constant $ c $ that makes the trinomial $ x^2 + 22x + c $ a perfect square, you can use the formula:

$$ c = \left(\frac{b}{2}\right)^2 $$

In this case, $ b = 22 $. So:

$$ c = \left(\frac{22}{2}\right)^2 = 11^2 = 121 $$

Therefore, the constant $ c $ that makes the trinomial a perfect square is $ c = 121 $.