Question

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

Answer 1

Hint: 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 \[{{p}^{2}}-5p+6<0.\] 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 \[{{p}^{2}}-5p+6\] in the form of \[\left( p-\alpha \right)\left( p-\beta \right)\] where \[\alpha \] and \[\beta \] are the zeroes of the polynomial. Then, the values of p at which \[{{p}^{2}}-5p+6\] is negative will lie between \[\alpha \] and \[\beta .\]


Complete step-by-step answer:
Before, solving the question, we will first determine what kind of polynomial is \[{{p}^{2}}-5p+6\] 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.
\[\Rightarrow {{p}^{2}}-5p+6<0.....\left( i \right)\]
Now, we will write \[{{p}^{2}}-5p+6\] as \[\left( p-\alpha \right)\left( p-\beta \right)\] where \[\alpha \] and \[\beta \] are the zeroes of this polynomial. We know that, if a quadratic equation of the form \[a{{x}^{2}}+bx+c\] is given, then its zeroes will be given by
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
In our case, a = 1, b = – 5 and c = 6. Thus, we will get,
\[p=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\left( 1 \right)\left( 6 \right)}}{2\left( 1 \right)}\]
\[\Rightarrow p=\dfrac{5\pm \sqrt{25-24}}{2}\]
\[\Rightarrow p=\dfrac{5\pm \sqrt{1}}{2}\]
\[\Rightarrow p=\dfrac{5\pm 1}{2}\]
\[\Rightarrow \alpha =\dfrac{5+1}{2}=\dfrac{6}{2}=3\]
\[\Rightarrow \beta =\dfrac{5-1}{2}=\dfrac{4}{2}=2\]
Thus, we can write \[{{p}^{2}}-5p+6\] as \[\left( p-2 \right)\left( p-3 \right).\] Now, we will put this value in (i). Thus, we will get,
\[\left( p-2 \right)\left( p-3 \right)<0\]
Now, if there is an inequality of the form \[\left( x-a \right)\left( x-b \right)<0\] then it will satisfy the values lying between a and b. Thus, the solution of the above inequality is
\[2 < p < 3\]
Hence, option (b) is the right answer.

Note:The answer of this question will be \[2