Question

Consider a quadratic equation 2x^(2)-mx+(m+1)=0.
If the roots of the equations are alpha and 2(alpha)+1, find the possible value(s) of m.

Answer 1

\(\alpha \) and \(2\alpha +1\)?

Answer 2

yes

Answer 3

for the equation:
ax^2+bx+c=0
sum of roots r_1+r_2=-b/a
product of roots: r_1*r_2=c/a

Answer 4

r_1+r_2???

Answer 5

sum of roots is \(-\frac{b}{a}\) in your case
\[\alpha +2\alpha +1=\frac{m}{2}\]
product of the roots is \(\frac{c}{a}\) you get \(\alpha (2\alpha +1)=\frac{m+1}{2}\)

Answer 6

r_1 is 1º root, r_2 2º root

Answer 7

by which i mean (and @myko means) the sum of the roots of
\[ax^2+bx+c=0\]

Answer 8

you get myko is typing a reply…

Type your reply
\[3\alpha +1=\frac{m}{2}\] or
\[m=6\alpha +2\]
\[\alpha^2+\alpha =\frac{m+1}{2}\] or
\[2\alpha ^2+2\alpha -1=m\]

Answer 9

set them equal and solve for \(m\) in terms of \(\alpha\)

Answer 10

\[3\alpha+1=m\]
\[\alpha=\frac{ m-1 }{ 3 }\] ---(1)
\[2\alpha ^{2}+\alpha=m+1\]---(2)

Sub. (1) into (2),we have