Question 158062
<font size = 6 color = "red"><b>Edwin's solution:</b></font>

Find the value for k that will make 4x^2+6.4x +k a perfect square.
0 solutions
<pre><font size = 4 color = "indigo"><b>
{{{4x^2+6.4x +k}}}

If it is to factor as a perfect square its 
factorizaton must be:

{{{(2x+sqrt(k))^2}}}

So 

{{{4x^2+6.4x +k = (2x+sqrt(k))^2}}}

{{{4x^2+6.4x +k = (2x+sqrt(k))(2x+sqrt(k))}}}

{{{4x^2+6.4x +k = (2x)^2+2x*sqrt(k)+2xsqrt(k)+ (sqrt(k))^2)}}}

{{{4x^2+6.4x +k = (2x)^2+4x*sqrt(k)+(sqrt(k))^2)}}}

{{{4x^2+6.4x +k = 4x^2+4x*sqrt(k)+k}}}

Since the first and last terms on each sides are equal, the
middle terms must also be equal. So we set them equal:

{{{6.4x = 4x*sqrt(k)}}}

Divide both sides by x:

{{{6.4 = 4sqrt(k)}}}

Divide both sides by 4:

{{{6.4/4 = sqrt(k)}}}

{{{1.6 = sqrt(k)}}}

Square both sides:

{{{1.6^2 = (sqrt(k))^2}}}

{{{2.56 = k}}}

Edwin</pre>