Quadratic equation with one root in $[0,1]$ and other root in $[1,\infty]$

Question

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?

Answer 1

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.

Answer 2

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