find a positive value for k such that the equation
xsquared + 2kx + 3(k+1) = 0
has exactly one real-number solution. Give two forms for your answer: one involving a radical and the other a calculator approximation rounded to two decimal places.
Is there an easier way to type exponents? thanks
xsquared + 2kx + 3(k+1) = 0
has exactly one real-number solution. Give two forms for your answer: one involving a radical and the other a calculator approximation rounded to two decimal places.
Is there an easier way to type exponents? thanks