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For what real values of c is \(4x^2+14x+c\) the square of a binomial?

Answer is NOT 49

 Dec 30, 2022
edited by Keihaku  Dec 30, 2022
 #1
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For it to be the square of a binomial, it has to be in the form:

 

\((r + s)^2 = 4x^2 + 14x + c\)

We find that the first term squared = 4x^2

\(r^2 = 4x^2\)

\(r = 2x\)

 

\((2x + s)^2 = 4x^2 + 14x + c\)

 

We know the second term is in the form:

\(4(x)(r) = 14x\)

\(4r = 14\)

\(r = 3.5\)

 

So our binomial is in the form:

 

\((2x+3.5)(2x+3.5) \)

So our constant "c" would be:

\(c = 3.5^2 \)

\(c = 12.25\)

 

(If this answer is wrong please tell me, as I would be more than happy to fix it)

 Dec 30, 2022
 #2
avatar+454 
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Its correct, thank you so much!

Keihaku  Dec 30, 2022

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