Question 766122
quadratic equation whose roots are twice of the given equation
ax^2+bx+c=0

is

ax^2+2bx+4c=0

You can derive using following-
Roots for the equation ax^2+bx+c=0 are given by

X1 = [-b + sqrt(b^2-4ac)]/2a
X2 = [-b - sqrt(b^2-4ac)]/2a

The new roots are twice of these roots
Y1 = 2X1 = 2*[-b + sqrt(b^2-4ac)]/2a
Y2 = 2X2 = 2*[-b - sqrt(b^2-4ac)]/2a

or

you can derive the equation by resolving below-

(x - Y1)*(x - Y2) = 0
(x - [-b + sqrt(b^2-4ac)]/a)*(x - [-b - sqrt(b^2-4ac)]/a) = 0