SOLUTION: Give the quadratic equation whose roots are twice as large as the roots of the given equation: ax^2+bx+c=0 a is not equal to 0.

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Question 766122: Give the quadratic equation whose roots are twice as large as the roots of the given equation: ax^2+bx+c=0 a is not equal to 0.
Found 2 solutions by subudear, josgarithmetic:
Answer by subudear(62) About Me  (Show Source):
You can put this solution on YOUR website!
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 = 2/2a
Y2 = 2X2 = 2/2a
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

Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
Highly believing that 'a' will not be affected, an attempt at guessing is tried, not really the best approach:
ax^2+2b+2c=0
roots be +x=%28-2b%2B-+2%2Asqrt%28b%5E2-a%2A2%2Ac%29%29%2F%282a%29
But seems to fail. Looking as 4c as replacement for c will be better, try:

ax%5E2%2B2b%2B4c=0
x=%28-2b%2B-+sqrt%284b%5E2-4%2Aa%2A4c%29%29%2F%282a%29
x=%28-2b%2B-+2%2Asqrt%28b%5E2-4ac%29%29%2F%282a%29
x=%28-b%2B-+sqrt%28b%5E2-4ac%29%29%2Fa, yes, this root pair is twice what we normally have in our solution formula.

COMPARE:
ax%5E2%2Bbx%2Bc=0
x=%28-b%2B-+sqrt%28b%5E2-4ac%29%29%2F%282a%29
versus
ax%5E2%2B2b%2B4c=0
x=%28-b%2B-+sqrt%28b%5E2-4ac%29%29%2F%28a%29