Find the value of $a$ such that at least one root of the equation $f(x)=x^2 - (a-3)x + a =0$ is greater than $2$.

Question

As the title says. To solve this problem I took two cases and solved them separately:
1. when the $x$ coordinate of the vertex is greater than $2$ and $f(2)>0$;
2. when $f(2)<0$.

However I wasn't able to find the answer as the two conditions have no common solution (obviously they won't have because in one case $f(2)>0$ and in the other one $f(2)<0$.

The inequalities which I got after solving for the two cases separately were:
1. $a<10$ and $a<7$
2. $10 < a$

And help would be appreciated. :)

Answer 1

The roots of that equation are $$ \frac{(a-3) \pm \sqrt{a^2-6a+9-4a}}{2} $$ so specifically we need $$ \frac{(a-3) + \sqrt{a^2-6a+9-4a}}{2} > 2 \implies \sqrt{a^2-10a+9} > 7 - a $$

Here's where I made a mistake at first. If $a < 7$ then we have

$$ a^2 - 10a + 9 > 49 - 14a + a^2 \implies a > 10 $$ so that this route gets us no solutions, now if $a > 7$ we just need to have $\sqrt{a^2 - 10a + 9}$ to be real which mean $a^2 - 10a + 9 > 0$. Notice that the roots to this second polynomial are $$ \frac{10 \pm \sqrt{100 - 36}}{2} = \frac{10 \pm 8}{2} = 9,1 $$ Now since we have the restriction that $a > 7$ we need to see whether the polynomial is positive or negative in $[7,9]$ and likewise from $[9,\infty)$. Plugging in $a=8$ we see that $$ 8^2-10(8)+9 = -16 + 9 < 0 $$ so $a \not\in [7,9]$. Now checking $a=10$ we have $$ 10^2 -10(10) + 9 = 9 > 0 $$ so that indeed $a > 9$.

Now we need to do something similar for the other root $$ \sqrt{a^2-10a+9} < a - 7 $$ Now if $a < 7$ this doesn't make sense since a square root can only be positive. Otherwise if $a > 7$ then we have $$ a^2 - 10a + 9 < a^2 - 14a + 49 \implies 4a < 40 \implies a < 10 $$ as long as $a^2 - 10a + 9 \ge 0 \implies a \ge 9$

So our other root gives us a solution if $a \in [9,10)$. Now combining these we get that $a \in [9, \infty)$.

Sorry for the issues before

Answer 2

$x^2-(a-3)+a=0$

$\therefore \alpha, \beta=\large\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{(a-3)\pm \sqrt{(a-3)^2-4a}}{2}=\frac{(a-3)\pm \sqrt{(a-1)(a-9)}}{2}$

$$\therefore\large \frac{(a-3)\pm \sqrt{(a-1)(a-9)}}{2}>2$$

$$\therefore(a-3)\pm \sqrt{(a-1)(a-9)}>4$$

$$\therefore \pm \sqrt{(a-1)(a-9)} > 4-a+3$$

$$\therefore (a-1)(a-9) > (7-a)^2$$

$$\therefore (a-1)(a-9) \geq 0$$

For this to be true either both brackets on LHS should be positive OR both negative.

$\therefore a \geq9$ OR $a \leq1$

EDIT: substitute $a=9$ and $a=1$ in the original equation and we see that $a=9$ satisfies the constraint while $a=1$ does not, so answer is $a \geq9$