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For what real values of $c$ is $x^2 - 8x - 14x + c$ the square of a binomial?

 May 27, 2024

Best Answer 

 #1
avatar+806 
+1

First, let's combine some like terms!

\(x^2 - 22c + c\)

 

In order for this to be the square of a binomial, we need c to be the square of half the value of -22. 

 

Solving this, we get c = 121. 

 

Indeed, if we have \(x^2-22x+121=(x-11)^2\)

 

So we have c = 121. 

 

Thanks! :)

 May 27, 2024
 #1
avatar+806 
+1
Best Answer

First, let's combine some like terms!

\(x^2 - 22c + c\)

 

In order for this to be the square of a binomial, we need c to be the square of half the value of -22. 

 

Solving this, we get c = 121. 

 

Indeed, if we have \(x^2-22x+121=(x-11)^2\)

 

So we have c = 121. 

 

Thanks! :)

NotThatSmart May 27, 2024

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