Solveeit Logo
Login

Question

Question: For which values of p is \[{{p}^{2}}-5p+6\] negative? (a) p < 0 (b) 2 < p < 3 (c) p > 3 (d) ...

For which values of p is p2−5p+6{{p}^{2}}-5p+6 negative?
(a) p < 0
(b) 2 < p < 3
(c) p > 3
(d) p < 2

Explanation

Solution

To solve the above question, we will first determine what kind of polynomial is P(p) and how many zeroes it will have. Then we will have to solve p2−5p+6<0.{{p}^{2}}-5p+6<0. For this, we will have to first find the values of p at which the left-hand side is zero. We will do this by the method of the quadratic formula. After doing this, we will write p2−5p+6{{p}^{2}}-5p+6 in the form of (p−α)(p−β)\left( p-\alpha \right)\left( p-\beta \right) where α\alpha and β\beta are the zeroes of the polynomial. Then, the values of p at which p2−5p+6{{p}^{2}}-5p+6 is negative will lie between α\alpha and β.\beta .

Complete step-by-step answer :
Before, solving the question, we will first determine what kind of polynomial is p2−5p+6{{p}^{2}}-5p+6 and how many zeroes it will have. As the highest power of p is 2, it is a quadratic polynomial in p and it will have two zeroes. Now, it is given that P(p) should be negative i.e. P(p) < 0.
⇒p2−5p+6<0.....(i)\Rightarrow {{p}^{2}}-5p+6<0.....\left( i \right)
Now, we will write p2−5p+6{{p}^{2}}-5p+6 as (p−α)(p−β)\left( p-\alpha \right)\left( p-\beta \right) where α\alpha and β\beta are the zeroes of this polynomial. We know that, if a quadratic equation of the form ax2+bx+ca{{x}^{2}}+bx+c is given, then its zeroes will be given by
x=−b±b2−4ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}
In our case, a = 1, b = – 5 and c = 6. Thus, we will get,
p=−(−5)±(−5)2−4(1)(6)2(1)p=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\left( 1 \right)\left( 6 \right)}}{2\left( 1 \right)}
⇒p=5±25−242\Rightarrow p=\dfrac{5\pm \sqrt{25-24}}{2}
⇒p=5±12\Rightarrow p=\dfrac{5\pm \sqrt{1}}{2}
⇒p=5±12\Rightarrow p=\dfrac{5\pm 1}{2}
⇒α=5+12=62=3\Rightarrow \alpha =\dfrac{5+1}{2}=\dfrac{6}{2}=3
⇒β=5−12=42=2\Rightarrow \beta =\dfrac{5-1}{2}=\dfrac{4}{2}=2
Thus, we can write p2−5p+6{{p}^{2}}-5p+6 as (p−2)(p−3).\left( p-2 \right)\left( p-3 \right). Now, we will put this value in (i). Thus, we will get,
(p−2)(p−3)<0\left( p-2 \right)\left( p-3 \right)<0
Now, if there is an inequality of the form (x−a)(x−b)<0\left( x-a \right)\left( x-b \right)<0 then it will satisfy the values lying between a and b. Thus, the solution of the above inequality is
2<p<32 < p < 3
Hence, option (b) is the right answer.

Note :The answer of this question will be [2