If a quadratic equation has equal roots, then shouldn't the answer you get be the same numbers? REPOST

the determinant = (m+4)^2 - 4(4m+1) = 0 to have equal roots
so m^2 +8m + 16 - 16m-4 = 0 so m^2 - 8m + 12 = 0
solve this, m = 2 or 6

If X^2 + (m+4)X + (4m+1) = 0 has equal roots we can write:

X^2 + (m+4)X + (4m+1) = (X + a)^2
where a is the double root of the equation.

(X + a)^2 = X^2 + 2aX + a^2

we now identify both equations to get:

m+4 = 2a and 4m+1 = a^2

Now we show that m^2 - 8m + 12 = 0

we have 2a = m+ 4 we square both sides to get:

4a^2 = m^2 + 8m + 16
divide by 4 to get
a^2 = (m^2 + 8m + 16)/4
but we have a^2 = 4m + 1

therefore :

m^2 + 8m + 16 = 4(4m + 1) <== equated and multiplied by 4
this yields

m^2 - 8m +12 = 0
This equation has m = 6 and m = 2 as solution

we go back and substitute for m to find which m yields the root for the original equation.

2a = m + 4 => a = 3 or a = 5 corresponding to m = 2 and m = 6 respectively
a^2 = 4m + 1 => a = sqrt(4m + 1) = sqrt(9) = 3 or m =sqrt(25) = 5

So the equation X^2 + (m+4)X + (4m+1) = 0 has two double roots one double root equal to 3 which corresponds to m = 2 and another double root of 5 which corresponds to m = 6