Find the values of $a$ for which $x^2-ax+2=0$ has one root in $[0,1]$ and other root in $[1,\infty]$.

The twoo rots are $$\frac{a\pm\sqrt{a^2-8}}{2}$$

The smaller root should be less than $1$.

So $$a-\sqrt{a^2-8}\le 2$$ $$a-2\le\sqrt{a^2-8}$$ $$a^2+4-4a\le a^2-8$$ $$a\ge 3$$

How will I find the upper bound for $a$? And what is the general approach to solve such problems where the roots are constrained between two values?

  • $\begingroup$ $a-\sqrt{a^2-8}\le 2$. $\endgroup$ – André Nicolas Mar 4 '16 at 7:30
  • $\begingroup$ Shouldn't your first inequality be divided by 2 (so that it is less than 2 and not 1)?! $\endgroup$ – user103828 Mar 4 '16 at 7:30
  • $\begingroup$ Regardless, I think your strategy is correct. I don't see a problem even if you get $a \geq 9/2$... now try plugging in the reverse inequality and the other sign and see what bounds you get. $\endgroup$ – user103828 Mar 4 '16 at 7:34
  • $\begingroup$ Oops. My mistake.... $\endgroup$ – Aditya Dev Mar 4 '16 at 7:38
  • $\begingroup$ Caution, $a<b$ doesn't imply $a^2<b^2$. $\endgroup$ – Yves Daoust Mar 4 '16 at 7:54

First the two roots need to exist, then $$a^2>8.$$

Then the two conditions are

$$a-\sqrt{a^2-8}\le2,\\a+\sqrt{a^2-8}\ge2,$$ or $$a-2,2-a\le\sqrt{a^2-8}.$$

This is equivalent to

$$(a-2)^2\le a^2-8,\\12\le 4a.$$

This condition is stronger than the first one.


For the upper bound you can use the rule:

Consider $ax^2 + bx + c = 0$ Multiplication of the roots are equal to $\frac{c}{a}$

That is, if $r_1*r_2 = 2$ in this question.

If $0<r_1<1$, then $r_2 = \frac{2}{r1} > 1$


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