Question

Consider the quadratic equation
m^2 x^2 + 2(2m-5)x + 8 = 0

a) If 2 is a root of the equation, find the possible values of m (there are four possible values)

b) For each value of m, find the other root

I have actually calculated the answer m=1 or m=-3 for (a), but I was told that there are four possible values of m. I was stuck on this part.

Answer 1

since 2 is a root,
4m^2 + 2(2m-5)(2) + 8 = 0
4m^2 + 8m - 20 + 8 = 0
4m^2 + 8m - 12 =
m^2 + 2m - 3 = 0
(m+3)(m-1) = 0
m = -3, or m = 1

if m = -3, we have
9x^2 -22x + 8 = 0
(x-2)(9x - 4) = 0
x = 2 (our given) or x = 4/9

if m = 1 , we have
x^2 -6x + 8 = 0
(x-2)(x-4) = 0
x=2 or x=4

Like you, I only get 2 values for m
and each value of m yields 2 answers. (one doubling up)

I also approached the problem using the sum and product of roots property.
let the roots be a and b, (we have a quadratic in x)
but we know one of them let b = 2

sum of roots = a+2
product of roots = 2a

sum of roots = -2(2m-5)/m^2
a+2 = (-4m + 10)/m^2 - 2
a = (-4m +10 - 2m^2)/m^2

product of roots = 8/m^2
2a = 8/m^2
a = 4/m^2

4/m^2 = (-4m + 10 - 2m^2)/m^2
4 = -4m + 10 - 2m^2
2m^2 + 4m - 6 = 0
m^2 + 2m - 3 = 0
(m+3)(m-1) = 0
m = -3 or m = 1

same as above, and only 2 values of m