Condition for roots to lie in certain intervals

Question

The set of values of $p$ such that both the roots of the equation $$f(x)=(p−5)x^2−2px+(p−4)=0$$ are positive and one of the roots is less than $2$ and the other root lies between $2$ & $3$ is ?

Are these conditions sufficient?Please check.

$f(2).f(3)<0$

$(p-5)f(2)<0$

$(p-5)f(0)>0$

$\frac{p}{p-5}<0$

P.S:Also tell me how to reduce the conditions to be checked for if possible.

Answer 1

Using the quadratic formula:

$$x = \frac{1}{2}\cdot \frac{2p\pm2\sqrt{9p-20}}{p-5}$$

We want the smaller root $x_1$ to satisfy $0<x_1<2$, so:

$$ 0<\frac{2p-2\sqrt{9p-20}}{2(p-5)}<2$$ $$\Rightarrow p\in A =(4,5)\cup (5,\infty)$$

And we want the bigger root $x_2$ to satisfy $2<x_2<3$, i.e.

$$2<\frac{2p+2\sqrt{9p-20}}{2(p-5)}<3$$

$$\Rightarrow p\in B=\bigg(\frac{49}{4},24\bigg)$$

So to satisfy both conditions, we must have

$$p\in A\cap B = \bigg(\frac{49}{4},24\bigg)$$

Answer 2

To make this easier for us, let's split this up into two cases:

1. $p-5 < 0$ means $p < 5$. In this case, the quadratic curves downwards because the leading term is negative., meaning before the first root and after the second root is negative, but in between is positive.
2. $p-5 > 0$ means $p > 5$. In this case, the quadratic curves upwards because the leading term is positive, meaning before the first root and after the second root is positive, but in between is negative.
3. $p-5 = 0$ is impossible because then, the leading term is $0$ and this becomes a line that has, at most, one root, but the problem calls for two.

Case 1:

Both roots are positive, so $0$ is before the first root, meaning $f(0) > 0$. $f(0)=p-4$, so $p-4 < 0$, meaning $p < 4$.

Since $2$ is between the roots, $f(2) > 0$. This means $(p-5)(4)-2p(2)+(p-4)=4p-20-4p+p-4=p-24 > 0$. Thus, $p > 24$. However, $p < 5 < 24$, so this is a contradiction.

Case 2:

Both roots are positive, so $0$ is before the first root, meaning $f(0) > 0$. $f(0)=p-4$, so $p-4 > 0$, meaning $p > 4$. However, we already know that $p > 5 > 4$, so this does not help us.

Since $2$ is between the roots, $f(2) < 0$. This means $(p-5)(4)-2p(2)+(p-4)=4p-20-4p+p-4=p-24 < 0$. Thus, $p < 24$.

Since $3$ is after the second root, $f(3) > 0$. This means $(p-5)(9)-2p(3)+(p-4)=9p-45-6p+p-4=4p-49 > 0$, meaning $p > \frac{49}{4}$.

Thus, if we combine $p < 24$ with $p > \frac{49}{4}$, we get $\frac{49}{4} < p < 24$ for our answer from this case. Since the rest of the cases led to a contradiction, this is our answer to the whole problem.