Hi Guest!
If \(\text{2n is a root of the equation in x, }\) \(x^2-2mx+2n=0, \text{then find m-n} \)
\(\text{Suppose the second root is t}\)
\(\text{Then by Vietas' formulae:}\)
\(2n+t=2m\) (1) (Sum of the roots is -b/a)
\(\text{Rearranging (1) to get:}\)
\(m-n=\frac{t}{2}\) (2)
\(\text{Now,}\) \(\text{by Vietas' formulae:}\)
\(2nt=2n \) (The product of the roots is c/a)
Solving this for t:
\(2nt-2n=0 \iff 2n(t-1)=0\) \(\text { So n=0 (which is a root), or t=1 (which is the second root).}\)
\(\text { Therefore, by (2): }\)
\(m-n=\dfrac{t}{2}=\dfrac{1}{2}\)
Hope this helps!
