Question #efd42

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Answer 1 · 2016-10-26T16:44:59

Given quadratic equation of m is

$2m^2-16m+8=0$

$=>m^2-8m+4=0$

It being a quasratic equation it will have only two values that satisfy the equation. Let these are a and b.

So $(m-a)(m-b)=0$ should be the qudratic equation.

Comparing these two equations
we have

$(m-a)(m-b)=m^2-8m+4$

$m^2-(a+b)m+ab=m^2-8m+4$

Comparing LHS and RHS we get

$a+b=8$

~~~~~~~~~~~~~~~~~~~~~~~~~~±~
For any quadratic equation

$ax^2+bx+c=0$ ,

if $alpha and beta$ are two values of x that satisfy the equation then

$alpha+beta=-b/a and alphabeta=c/a$