3. Find the range of values of m if $\displaystyle x^2+x+1=m(x+2)$ has real and distinct roots.

This is the same as #2, but this time, since we want real and distinct roots, the discriminant has to be positive, or $\displaystyle b^2 - 4ac > 0$. So

$\displaystyle \begin{aligned}

x^2 + x + 1 &= m(x + 2) \\

x^2 + x + 1 &= mx + 2m \\

x^2 + x - mx + 1 - 2m &= 0 \\

x^2 + (1 - m)x + (1 - 2m) &= 0 \\

\end{aligned}$

a = 1

b = 1 - m

c = 1 - 2m

$\displaystyle \begin{aligned}

b^2 - 4ac &> 0 \\

(1 - m)^2 - 4(1)(1 - 2m) &> 0 \\

1 - 2m + m^2 - 4 + 8m &> 0 \\

m^2 + 6m - 3 &> 0

\end{aligned}$

I want to find the zeros of the polynomial $\displaystyle m^2 + 6m - 3$. Let's pretend for a moment that it equals 0 and solve for m:

$\displaystyle \begin{aligned}

m^2 + 6m - 3 &= 0 \\

m^2 + 6m &= 3 \\

m^2 + 6m + 9 &= 3 + 9 \\

(m + 3)^2 &= 12 \\

m + 3 &= \pm\sqrt{12} \\

m &= -3 \pm 2\sqrt{3} \\

\end{aligned}$

Make a sign chart. Draw a number line and the critical points (the zeros we found earlier):

Code:

---------+-------------------+---------
-6.46 0.46

($\displaystyle 0.46 \approx -3 + 2\sqrt{3}$ and $\displaystyle -6.46 \approx -3 - 2\sqrt{3}$)

You have 3 intervals to test:

$\displaystyle (-\infty,\;-3 - 2\sqrt{3})$

$\displaystyle (-3 - 2\sqrt{3},\;-3 + 2\sqrt{3})$

$\displaystyle (-3 + 2\sqrt{3}, \infty)$

From each interval, pick a number inside it to test into the inequality $\displaystyle m^2 + 6m - 3 > 0$, in order to determine the sign. The signs of the polynomial for each interval is as follows:

Code:

---------+-------------------+---------
pos -6.46 neg 0.46 pos

So the answer is $\displaystyle m<-3-2\sqrt{3},m>-3+2\sqrt{3}$.

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