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

 Dec 26, 2020
edited by Guest  Dec 26, 2020
 #1
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+1

(2x  + 3.5)(2x  + 3.5  )

 

(2x+3.5)2 =    4 x2 +14x + 12.25

 Dec 26, 2020
 #2
avatar+112863 
+1

I'd do it like this

 

\(4x^2 + 14x + c\\ =4(x^2+\frac{14}{4}x+\frac{c}{4})\\ =4(x^2+\frac{7}{2}x+\frac{c}{4})\\\)

 

now if this is a perfect square then:

 

   

\(\frac{c}{4}=[\frac{7}{2}\color{red}\div2\color{black}]\color{red}^2\\ \frac{c}{4}=[\frac{7}{4}]\color{red}^2\\ \frac{c}{4}=\frac{49}{16}\\ c=\frac{49}{4}=12\frac{1}{4}\)

 

 

 

 

LaTex:

\frac{c}{4}=[\frac{7}{2}\color{red}\div2\color{black}]\color{red}^2\\
\frac{c}{4}=[\frac{7}{4}]\color{red}^2\\
\frac{c}{4}=\frac{49}{16}\\
c=\frac{49}{4}=12\frac{1}{4}

 Dec 28, 2020

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