Am I doing Vieta's Formulas correctly?

Question

I've been given this exercise: $x^2 - (m+3)x + m + 2$, I'm supposed to check for which real values of the parameter $m$ can be used for this to work:

$1/x_1 + 1/x_2 > 1/2$ (I multiply both sides by 2x1x2) and get the following:

$2(x_1 + x_2) -x_1x_2 > 0$

which results to $m > -4$ using Vieta's formulas

Afterwards I have another argument, $x_1^2 + x_2^2 < 5$

After solving I get that $m$ is in the interval $(-4,0)$

My book tells me the final result for possible $M$ solutions is in the interval $(-2,0)$.

What am I doing wrong?

Answer 1

You can't multiply by $x_1x_2$ since you don't know if it's a positive or negative quantity (remember the sign of the inequality would have to swap if it were negative, and stay the same otherwise).

Remember what Viete's formulae tell you, that $x_1+x_2 = m+3$ and that $x_1x_2 = m+2$. You can use these if you simplify the left-hand side: $$\frac 1{x_1} + \frac1{x_2} = \frac{x_1+x_2}{x_1x_2} = \frac{m+3}{m+2},$$ so you want to ensure that $m$ is such that $$\frac{m+3}{m+2}>\frac12.$$ We can't multiply throughout by $m+2$ since we don't know its sign. We can multiply by $(m+2)^2$, this is surely non-negative. This gives us $$(m+3)(m+2)>\frac12(m+2)^2$$ which simplifies to $$(m+2)(m+4)>0.$$ A product of two numbers is $>0$ either if they are both $>0$, or if they are both $<0$.

In the first case (when $m+2$ and $m+4$ are both positive), we have $m>-2$ and $m>-4$, which is simply equivalent to saying $m>-2$.

In the second case (when they are both negative), we have $m<-2$ and $m<-4$, which is the same as saying that $m<-4$.

So in summary, your condition is equivalent to saying that $$\boxed{\text{$m<-4$ or $m>-2$}}.$$

Answer 2

In inequalities, it's usually a good idea to combine the fractions if you are not sure about their signs. For examples see here and here.

Now $$\frac{1}{x_1} + \frac{1}{x_2} > \frac 12 \iff \frac{2x_2+2x_1-x_1 x_2}{2x_1 x_2} >0 \\ \iff \frac{2(m+3)-m-2}{2(m+2)} = \frac{m+4}{2(m+2)}>0 \iff (m+2)(m+4) > 0 \\\iff m \in (-\infty, -4)\cup (-2, \infty)$$ And $$x_1^2+x_2^2 < 5 \iff (x_1+x_2)^2-2x_1 x_2 < 5 \\\iff (m+3)^2-2(m+2)-5 = m(m+4) < 0\\ \iff -4<m<0 $$

Therefore $-2<m<0$.