What is the set of of values of $p$ for which one root is less than $1$ and the other is greater than $2$ ?
A) $ (7/3,\infty) $
B) $ (-\infty,7/3) $
C) $ x \in R $
D) $ None $
Please tell me whats wrong with each of my approach :
My Approach 1 :
$ \alpha \in (-\infty,1) $ & $ \beta \in (2,\infty) $
$\implies -\infty + 2 < \alpha + \beta < 1+\infty $
$\implies \alpha + \beta \in R $ $\implies p \in R $
Now D > 0 , for two distinct roots . => ( p – 3 ) ( p – 5 ) > 0
$ \implies p \in ( -\infty , 3 ) \cup ( 5 , \infty ) $
My Approach 2:
$ \alpha \in (-\infty,1) $ & $ \beta \in (2,\infty) $
from equation
$ \alpha + \beta = 2p \implies \alpha = 2p – \beta $
$\alpha \cdot \beta = 8p – 15 $
$\implies ( 2p – \beta ) \cdot ( \beta ) = 8p – 15 $
$\implies p = {{\beta^2 – 15}\over {2\beta – 8}} $
Now how do I find p from this using b > 2 ?!!
The answer is (B)
Best Answer
There are a number of mistakes in your attempt, so I'm just going to show you a new method.
Let $y(x)=x^2-2px+8p-15$ . Then our quadratic equation is simply $y=0$. You probably know that if you plot the graph of $y$ against $x$, it will be an upward opening parabola. The roots of $y=0$ will simply be the abcissa of the points where the parabola cuts the X-axis.
Now for two real roots, $D>0$ so we have $4p^2>4(8p-15)\tag{i}$
Also if one root is less than 1 and the other is greater than 2 then you can imagine that the parabola will go below the X-axis for the entire interval $[1,2]$
Mathematically, we write this as $y(1)<0$ and $y(2)<0$ which gives us two more inequalities in $p$:
$$6p-14<0\tag{ii}$$ $$4p-11<0\tag{iii}$$ If you find the solution sets of the three inequalities obtained their intersection will be the final solution.