We have the equation $x^2+ax+2a+1=0$ which has real roots $x_1$ and $x_2$ and a is a parameter. I need to answer to the following questions:
Find all values of a for which $x_1=(a-1)x_2$.
For which values of a the roots $x_1$ and $x_2$ are positive?
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chen h.
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Hint:
$$\begin{align} x_1+x_2&=-a\\ x_1 x_2 &= 2a+1 \end{align}$$
Daniel R
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I know these formulae but I cant express $\frac{x_1}{x_2}$ in terms of a. – chen h. Nov 17 '14 at 11:03
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@chenh. Use the first expression to write $x_1$ in terms of $x_2$, then substitute it in $x_1=(a-1)x_2$. – Hakim Nov 17 '14 at 11:04
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And then? What do I do next?! – chen h. Nov 17 '14 at 11:05
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@chenh. You solve for $a$. – Hakim Nov 17 '14 at 11:05
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I got $x_2=-1$ is it true? – chen h. Nov 17 '14 at 11:27
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And then $a=0$. Is it correct? – chen h. Nov 17 '14 at 11:28
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1) $$x_1x_2=2a+1$$ $$\therefore x_2=\frac{2a+1}{x_1}$$ $$\therefore x_1=(a-1)\left(\frac{2a+1}{x_1}\right)$$ $$\therefore x_1^2=(a-1)(2a+1)$$ LHS is always greater than zero. $$\therefore (a-1)(2a+1)>0$$ From there, you will get the interval of $a$.
2) Just put RHS of $x_1+x_2$ and $x_1x_2$ greater than zero, you wll get the range of $a$.
Also, you have discriminant greater than zero. $$\therefore a^2-4(2a+1)>0$$ $$\therefore |a-4|>2\sqrt5$$
In the end, find the intersection of the range of values of $a$ from the answer and from the fact that discriminant is greater than zero.
Tejas
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