Question 1182830
The positive value of k which will make
4x2 - 4kx + 4k + 5 = 0 a perfect square trinomial is _____.


-- Please help me..
<pre>{{{matrix(1,3, 4x^2 - 4kx + 4k + 5, "=", 0)}}}
     {{{matrix(1,3, ax^2 + bx + c, "=", 0)}}} <==== Standard form of a quadratic
By EQUATING terms, we see that: {{{matrix(1,3, highlight(4)x^2 + highlight(- 4k)x + highlight(+ 4k + 5), "=", 0)}}}
                                {{{highlight(a)x^2}}}+  {{{highlight(b)x}}}+  {{{highlight(c)}}}  =  0
              Therefore, we get: a = 4
                                 b = - 4k
                                 c = + 4k + 5 

                                 {{{matrix(1,3, highlight(a)x^2 + highlight(b)x + highlight(c), "=", 0)}}} 
In order for the polynomial to be a PERFECT SQAURE, the discriminant (b<sup>2</sup> - 4ac) MUST = 0. 
                    We then get: {{{matrix(6,3, b^2 - 4ac, "=", 0, (- 4k)^2 - 4(4)(4k + 5), "=", 0, 16k^2 - 16(4k + 5), "=", 0, 16k^2 - 64k - 80, "=", 0, 16(k^2 - 4k - 5), "=", 16(0), k^2 - 4k - 5, "=", 0)}}}
                                        (k - 5)(k + 1) = 0
                                    k - 5 = 0    OR      k + 1 = 0
                                        k = 5    OR         k = - 1

Since a  POSITIVE value for k is required, the ONLY ANSWER is: {{{highlight_green(matrix(1,3, k, "=", 5))}}}</pre>