6
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ $a-\sqrt{a^2-8}\le 2$. $\endgroup$ – 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
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $r_2 = \frac{2}{r1} > 1$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.