Question 454646
Determine the parameter k so that the quadratic equation k x^2-16x+16=0 had exactly one real solution.
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From the quadratic formula the zeros are:
{{{(-b +- sqrt(b^2-4ac))/2a}}}
In order for there to be only one real zero, the discriminant 
{{{b^2-4ac}}} must equal zero.
In this case, a = k, so we have
16^2 - 4*k*16 = 0
16 = 4k
This gives k = 4
As a check, we plot the graph, and confirm that the vertex lies on the x-axis:
{{{graph(300,300,-6,6,-20,20,4x^2-16x+16)}}}