Question 262987
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what value of k will make 16x^2/9 (fraction) - kx + 36 a perfect square trinomial?
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        The post by @mananth gives incorrect answer to the problem's question.

        See my solution below for correct, complete and clear solution.



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I will give you / (show you) two ways to solve the problem, for your benefits.


            <U><B>S o l u t i o n   1 </B></U>


Your trinomial has a form  a^2*x^2 - kx + 36,  where  a^2 = {{{16/9}}},  a = {{{sqrt(16/9)}}} = {{{4/3}}}.


It will be a complete square if and only if 

    k = +/- {{{2a*6}}} = +/- {{{2*(4/3)*6}}} = +/- (2*4*2) = +/- 16.    <<<---===  <U>ANSWER</U>


            <U><B>S o l u t i o n   2 </B></U>


Consider the discriminant of the given polynomial 

    d = " b^2 - 4ac ",   where  b = -k,  a = {{{16/9}}},  c = 36.


So,  d = (-k)^2 - 4*(16/9)*36 = k^2 - 16^2.


The given polynomial is a perfect square if and only if the discriminant is zero

    k^2 = 16^2,  which  implies  k = +/- 16.


You get the same answer as in Solution 1 above.
</pre>

Solved correctly and completely, in clear and transparent form.