Quadratic question (1 Viewer)

[enigma_1]

How do you write worked solution to this?

For what values of m are the roots of x^2 +2x+3 = m(2x+1) real and positive?

Answer:

2<=m<3

EDIT: Sorry the I didn't write the answer properly. How do you do it?

 

[QZP]

This is a very complex question; I think you may have written it wrong (please check again). The answer is infact 2 <= m < 3

 

[RealiseNothing]

This is a very complex question; I think you may have written it wrong (please check again). The answer is infact 2 <= m < 3

it's pretty straight forward

 

[QZP]

Haha yeah my bad. Poor choice of words. I will edit this with solution. Hold on

After expanding and factorising,
x^2 + 2(1-m)x + (3-m) = 0

For roots to be real, discriminant >= 0
i.e. 4(1-2m+m^2) - 4(3-m) >= 0
4m^2 -4m -8 >= 0
4(m-2)(m+1) >= 0
Therefore m <= -1 or m >= 2

Now, for roots to be positive, sum of roots > 0
i.e. -4(1-m) > 0
m > 1

AND

product of roots > 0
i.e. 3-m > 0
m > 3

Therefore 1< m < 3

Now, (m <= -1 or m => 2) U (1 < m < 3) Someone comment on use of union here? It belongs to set theory so I don't know if its appropriate - just trying to find ways to be more elegant. @ OP basically find the intersection of the two inequalities
Yields 2 <= m < 3