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if 2n is a root of the equation in x, x^2-2mx+2n=0, then m-n=

 Jul 5, 2022
 #1
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m - n = -2.

 Jul 5, 2022
 #2
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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!

 

 
 Jul 5, 2022
 #3
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It helped me on a current project, thank you!

 

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 Jul 5, 2022

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