Question 707046
There will be one and only one real solution to a quadratic equation when the discriminant is zero. So let's start by putting the equation into standard form. Adding {{{16x^2}}} and 25 to each side:
{{{16x^2+10kx+25=0}}}
The discriminant, in general, is {{{b^2-4ac}}}. Our "a" is 16, "b" is 10k and "c" is 25. So our discriminant is:
{{{(10k)^2-3(16)(25)}}}
which simplifies to:
{{{100k^2-1200}}}
We want this to be zero so:
{{{100k^2-1200 = 0}}}
Now we solve for k. Dividing both sides by 100:
{{{k^2-12 = 0}}}
Adding 12:
{{{k^2 = 12}}}
So {{{k = sqrt(12)}}} or {{{k = -sqrt(12)}}}. These square roots simplify:
{{{k = sqrt(4*3)}}} or {{{k = -sqrt(4*3)}}}
{{{k = sqrt(4)*sqrt(3)}}} or {{{k = -sqrt(4)*sqrt(3)}}}
{{{k = 2*sqrt(3)}}} or {{{k = -2*sqrt(3)}}}
These are the only possible values for k which will lead to one and only one real solution for x.