Question 1182830
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It appears to me that the response from the other tutor is only useful for a student who already knows how to solve the problem.<br>
Assuming you posted the question because you don't know how to solve the problem, I will provide a response that I hope will help you learn how.<br>
For the general quadratic equation<br>
{{{y = ax^2+bx+c}}}<br>
the roots are given by the quadratic formula,<br>
{{{(-b+-sqrt(b^2-4ac))/2a}}}<br>
The radicand {{{b^2-4ac}}} is the discriminant -- it tells whether the equation has 0, 1, or 2 real zeros (roots).  If the discriminant is 0, then the equation has a single root; that means the quadratic is a perfect square trinomial.<br>
So you need the discriminant of the given trinomial to be 0.<br>
{{{y = (4)x^2 + (-4k)x + (4k+5)}}}<br>
In this trinomial, a=4, b=-4k, and c=4k+5.  The discriminant is<br>
{{{b^2-4ac = (-4k)^2-4(4)(4k+5) = 16k^2-64k-80}}}<br>
So the trinomial is a perfect square if the discriminant {{{16k^2-64k-80}}} is 0:<br>
{{{16k^2-64k-80=0}}}
{{{k^2-4k-5=0}}}
{{{(k-5)(k+1)=0}}}
{{{k=5}}} or {{{k=-1}}}<br>
ANSWER: the trinomial is a perfect square if k is either 5 or -1.<br>
CHECK:
k=5: {{{4x^2-20x+25 = (2x-5)^2}}}
k=-1: {{{4x^2-4k+1 = (2x-1)^2}}}<br>