Case C: No real roots bs 3 ns- 12. A real quadratic Q(P) that is concave down an

Question

Case C: No real roots bs 3 ns- 12. A real quadratic Q(P) that is concave down and that has no real roots must have two complex-conjugate roots given say by ri, r2 = m ± ib. 12.A Show that Q(P) =-a(P-ri) (P-T2) can be ex- panded out as rt Q(P)a(P-m)2+b2]. 12.B Verify that m represents the mirror-line of symmetry of this quadratic. (Recall that we know the mirror line satisfies m > 0 in our cases of interest.) : 12.C Find the sequence of algebraic substitutions ( translations and rescalings in the variables P and t) that reduces the differential equation to the normalized form dS =-(s2 + 1). 12.C Give an algebraic formula for P in terms of S. 12.D Give a formula for the new time variable in terms of the old time. of ds

Explanation / Answer

12. A.

Q (p)= -a [(p-r1)(p-r2)]

= -a [p2-pr1-pr2+r1r2 ]

= -a [p2-p (r2+r1)+ (m+ib )(m-ib)]

=-a [p2-p (m+ib+m-ib)+m2-i2b2]

=-a [p2-p (2m)+m2-(-b2)

=-a [(p-m)2+b2], hence proved

12.B.

A parabola represents the graph of a quadratic function, each parabola has a line of symmetry which is called the axis of symmetry. The equation of the axis of symmetry is always a real number a constant, it's equation is x=n where n is a real number. Now,

For this case of imaginary roots

Q (p)= -a [(p-r1)(p-r2)]= -ap2+a (r1+r2)p+r1r2

To find the axis of symmetry use the formula

x=-b/(2a) for an equation of parabola in the form ax2+bx+c

using this formula for Q (p) we get

x=-a (r2+r1)/(-2a)

=(m+ib+m-ib)/2

=2m/2

=m thus x=m is the line of symmetry,about which the mirror image of the imaginary roots exists.

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