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What makes this a perfect square trinomial? x^2+22x+? Is the ? 11
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because 11*2 is 22
11^2
Wait the answer is 11^2? But wouldn't that be 121?
\[(a+b)^2=a^2+2ab+b^2\] and here a = x, 2ab = 22x making b 11, and therefor you want \[x^2+22x+121=(x+11)^2\]\]
\[?=121\] yes
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11. yes
????
this is not an equation. the question was what should the constant be to make the expression a perfect square. the answer is the constant should be 121 so you will have a perfect square, namely \[(x+11)^2=x^2+22x+121\]
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