Polynomials Question

Question

One root of ax^2 + mx + n = 0 is three times the other. Show that 3m^2 - 16an = 0.

Answer 1

x = [-m +or- sqrt( m^2 - 4an)] /( 2a) by quadratic formula

since one solution is 3 times the other, positive branch = 3 * negative branch:

[-m + sqrt( m^2 - 4an)]/(2a) = 3[-m - sqrt( m^2 - 4an)]/(2a)

multiplies both sides by 2a

-m + sqrt( m^2 - 4an)] = 3[-m - sqrt( m^2 - 4an)]

distributes 3:

-m + sqrt(m^2 - 4an) = -3m - 3*sqrt(m^2 - 4an)

adds m+3*Sqrt(m^2-4an) to both sides:

4*sqrt(m^2 - 4an) = -2m

squares both sides:

16 (m^2 - 4an) = 4m^2

distributes 16:

16m^2 - 64an = 4m^2

subtracts 4m^2 from both sides:

12m^2 - 64an = 0

divides by 4:

3m^2 - 16an = 0

end of proof