+0  
 
0
49
4
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For what real values of \(c\) is \(4x^2+14x+c\) the square of a binomial?

 May 13, 2022
 #1
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0

Taking the discriminant, we get c = 49.

 May 13, 2022
 #2
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It' says it's wrong

Guest May 13, 2022
 #3
avatar+1810 
+1

We know that the bionomial must be \((2x + y)^2\), that way the 2 2x's multiply out to \(4x^2\)

 

Expanding the bionomial out as is gives us: \(4x^2+2xy+2xy+y^2\)

 

We want the 2 2xy's to sum to 14x, so we have the equation: \(4xy= 14x\). Solving for y, we find \(y = {7 \over2}\)

 

Now, we have to findsubsitute this in, and expand. 

 

Hint: The co-efficent will be \(y^2\)

 May 13, 2022
 #4
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But what's the answer

 May 14, 2022

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