How does the discriminant of the general second-degree equation help identify degenerate cases?

Oct 20, 2011 ... (a) Prove that the discriminant D = b2 - 4ac ∈ K has a square root. /. D in L and that we have L = K(. /. D). [Hint: use the quadratic formula ... The answer is statement 1 only, which would be selection A. A graph that meets the criteria expressed above would be: -x^2 + x - 1. We consider the two solutions in (2) one at a time. With the plus, we get x+ = −b+ b2 − 4ac. 2a. = −b+ b2 − 4ac. 2a......... −b− b2 − 4ac. −b− ... If B2 -4AC < 0, the conic is a circle, or an ellipse. If B2 - 4AC = 0, the conic is a parabola. Another way to classify conics has to do with ... b± b. 2 4ac. 2a. E= dV dx. F=q E. V = 1. 4. Q r. R. 2 =N a. 2 x. 2 =2 Dt c=6 R. = v R. Oct 2, 2022 ... Find all the roots of the solution, and compute the distance between successive roots to see that it is a constant. Mar 4, 2018 ... Explanation: · ⇒ · + · + · 1 · = · 0 · ← · in standard form. with a=2,b=3 and c=1. ⇒b2−4a ... Page 1. E = mc. 2 x = -b 소. / b2 - 4ac. 2a e iπ. +1 = 0 e x. = с. ∑ k=0 x n. /n! ∫ tanx dx = -ln(cosx). Jan 24, 2019 ... ... 2+bx+c=0 if and only if. x={-b± √{b2-4ac}}/{2a}. LaTeX Original \documentclass{article} \usepackage{amssymb, mathrsfs, enumitem} \usepackage ... (b) The point P = (1,−1,2) lies on both the paraboloid x2+y2 = z and the ... If a 6= 0 and b2 − 4ac > 0, then y has the form c1eαx + c2eβx where α ...