# limit (x1) as m->infinity ( x1 - the lowest root )

#### Vali

##### Member
I have the following equation: x^2 - 2(m+1)x + 3m + 1=0
Also, I know that x1 is the lowest root of this equation.
I need to solve lim (x1) as m->infinity
A. 1
B. 3/2
C. 0
D. -1/2
E. -1
I tried to solve the equation with the discriminant then to calculate the limit but didn't work.
Also, I think that because x1 is the lowest root and the function graphic is a parabola, I tink that -b/2a (the peak of parabola) > x1 but I don't see how this condition would help me.
Some ideas?
Thanks!

#### Opalg

##### MHB Oldtimer
Staff member
I have the following equation: x^2 - 2(m+1)x + 3m + 1=0
Also, I know that x1 is the lowest root of this equation.
I need to solve lim (x1) as m->infinity
A. 1
B. 3/2
C. 0
D. -1/2
E. -1
I tried to solve the equation with the discriminant then to calculate the limit but didn't work.
Also, I think that because x1 is the lowest root and the function graphic is a parabola, I tink that -b/2a (the peak of parabola) > x1 but I don't see how this condition would help me.
Some ideas?
Thanks!
You did right to start by solving the equation, and you probably found that the lower root is $x_1 = m+1 - \sqrt{m^2-m}$. The trick now is to make that into a fraction, multiplying and dividing by $m+1 + \sqrt{m^2-m}$ to get $$x_1 = \frac{\bigl( m+1 - \sqrt{m^2-m}\bigr)\bigl( m+1 + \sqrt{m^2-m}\bigr)}{m+1 + \sqrt{m^2-m}}.$$ Can you take it from there, to get the limit as $m\to\infty$?

#### Vali

##### Member
Yes, I replaced m with x because I usually work with x.
Thank you very much for your help

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#### Greg

##### Perseverance
Staff member
I need to solve lim (x1) as m->infinity
Hi Vali; do you mean "What number does x1 approach as m grows without bound"?

#### Vali

##### Member
Hi Vali; do you mean "What number does x1 approach as m grows without bound"?
Yes.Sorry if I didn;t use the correct words.

#### Greg

##### Perseverance
Staff member
Hey, no problem. The notation is unusual so I have asked for a clarification to benefit those who may not understand. Happy foruming!