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Suppose a real-world situation is modeled by a quadratic equation in the form of at^2+bt+c=0 where t represents time. What must be true about the equation if the problem situation only has one solution?

User Tinika
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Consider, the real-world situation is modeled by a quadratic equation in the form of at^2+bt+c= x, where x is some variable. then, x = a*t2 + b*t + c
or a*t2 + b*t + c = x
or a*t2 + b*t + c-x = 0 Now use the general quadratic solution, t =
\frac{b +/- \sqrt{ b^(2) -4a(c-x)} }{2a} But here, a cannot be equal to 0.

Also, sqrt{ b^{2} -4a(c-x)} }{2a} ≥ 0otherwise, t will be an imaginary number, and in the real world, time cannot be imaginary. Now, to find out a solution we restrict to the positive terms only nd neglect negative signs. Then , our solution is
t = -b / (2a)
User Ritesh Sinha
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