Question

can you please derice hiw ti get C/A > 0 and B/A > 0 from the quadratic equation.

astle ersity SOLUTION OF SECOND ORDER EQUATION Substituting into Equation Ai, we get: Eliminating x0eơt we get the characteristic equation So that the roots are: 2A 19

Answer 1

For a quadratic equation i.e polynomial of degree '2'. f(x) = ax^2 + bx + c = 0 rightarrow (1) The roots are of (1) then sum of the roots is x_1 + x_2 = -b/a Product of the roots is x_1, x_2 = c/a In the problem given sigma_1, sigma_2 are negative real no's i.e x_1, x_2 in my notation. = > x_1 times x_2 = Product = +ve (Because both roots are negative real no's) \r\n i.e. Product of two negative no's is always positive i.e > 0 = > Product of roots > 0 But product of roots = c/a = > C/a > 0 x_1 + x_2 = -(x_1 + x_2)(Because both negative real numbers) but x_1 + x_2 = -b/a = > - (x_1 + x_2) = -b/a = > x_1 + x_2 = b/a Also given, a > 0 & b > 0 Therefore b/a > 0