Quadratic Equation (M.C.)

D
x^2+bx+c=0
The two roots are[ -b ± √(b^2-4c) ]/2

x^2-2bx+4c=0
The two roots are [ 2b ± 2√(b^2-4c) ] / 2

consider -b and 2b
as those choices didn't provide summation or subtraction of b.
to fulfill 2b, z can only multiply by -2





so if z = [ -b + √(b^2-4c) ]/2
a root of x^2-2bx+4c=0 is [ 2b - 2√(b^2-4c) ] / 2

if z = [ -b - √(b^2-4c) ]/2
a root of x^2-2bx+4c=0 is [ 2b + 2√(b^2-4c) ] / 2

2015-05-28 21:43:03 補充：
i realized that ive got something wrong and 002 has a better solution

As Z is the root of the equation x²＋bx＋c＝0, so
Z²＋bZ＋c＝0
==> c＝-Z²－bZ
For the equation x²－2bx＋4c＝0, it becomes
x²－2bx－4(Z²＋bZ)＝0
==> x²－4Z²－2bx－4bZ＝0
==> (x＋2Z)(x－2Z)－2b(x＋2Z)＝0
==> (x＋2Z)(x－2Z－2b)＝0
==> x＝-2Z or 2Z＋2b

So the answer is (D).