Posted by Jayden Haddy on .
Consider the quadratic equation
m^2 x^2 + 2(2m5)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.

maths 
Reiny,
since 2 is a root,
4m^2 + 2(2m5)(2) + 8 = 0
4m^2 + 8m  20 + 8 = 0
4m^2 + 8m  12 =
m^2 + 2m  3 = 0
(m+3)(m1) = 0
m = 3, or m = 1
if m = 3, we have
9x^2 22x + 8 = 0
(x2)(9x  4) = 0
x = 2 (our given) or x = 4/9
if m = 1 , we have
x^2 6x + 8 = 0
(x2)(x4) = 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(2m5)/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)(m1) = 0
m = 3 or m = 1
same as above, and only 2 values of m