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?

Respuesta :

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 = [tex] \frac{b +/- \sqrt{ b^{2} -4a(c-x)} }{2a} [/tex] 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)
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