Answer:
The greatest possible value of the second root is 54
Step-by-step explanation:
Given quadratic equation is [tex]x^2-5mx+6m^2=0[/tex]
Now factorise the given quadratic equation :
[tex]x^2-3mx-2mx+6m^2=0[/tex]
x(x-3m)-2m(x-3m)=0
(x-2m)(x-3m) = 0
x-2m=0 or x-3m=0
∴ x = 2m or x = 3m .
Therefore 2m and 3m are the roots of the given quadratic equation.
Given one of the roots of the quadratic equation is 36
Let 2m=36
[tex]m=\frac{36}{2}[/tex]
[tex]=18[/tex]
Therefore m=18
Substitute m=12 in 2m we have 2(12)=24
Now let 3m=36
[tex]m=\frac{36}{3}[/tex]
[tex]=12[/tex]
Therefore m=12
Substitute m=18 in 3m we have 3(18)=54
Therefore the greatest possible value of the second root is 54
