Question

If p, q are the roots of the equation ${{x}^{2}}+px+q=0$ , then the value of p must be equal to
(a) 0,1
(b) 1
(c) 2
(d) 1,-2

Answer 1

Hint: We know that for a quadratic equation of the form $a{{x}^{2}}+bx+c=0$ , the sum of roots $=-\dfrac{b}{a}$ and the product of roots $=\dfrac{c}{a}$ . Using the information that p and q are the roots of the quadratic equation, we will have two equations to find the value of two variables p and q.

Complete step by step answer:
We know that the general quadratic equation is of the form $a{{x}^{2}}+bx+c=0$ . For this equation, we can say that
Sum of roots $=-\dfrac{b}{a}$
Product of roots $=\dfrac{c}{a}$
When we compare this general equation with the given equation, we get
$a=1,\text{ }b=p\text{ and }c=q.$
So now, we have
Sum of roots $=-\dfrac{b}{a}=-p$
But it is given that the roots are p and q.
So, $p+q=-p...\left( i \right)$
Also, we have
Product of roots $=\dfrac{c}{a}=q$
But we have p and q as roots of the quadratic equation.
So, $pq=q...\left( ii \right)$
Here, let us take two cases to solve equation (i) and (ii),
Case 1: $q\ne 0$
We can cancel the q term from both sides of equation (ii), and then we will get
p = 1.
Substituting the value of p in equation (i), we get
1 + q = -1
Rearranging the terms, we get
q = -2.
Hence, the value of p is 1.
Case 2: $q=0$
For this case, we will not get any useful information from equation (ii).
Substituting the value of q in equation (i), we get
p + 0 = -p
We can also write this as
2p = 0
Hence, we have p = 0.
Thus, the value of p is 0.
Hence, p can attain the values 0 and 1.
So, the correct answer is “Option a”.

Note: We must remember that we can only cancel terms from two sides of any equation only if the value of that term is not equal to zero. We can also solve this problem by forming the quadratic equation using the roots as $\left( x-p \right)\left( x-q \right)=0$ and then by comparison with the given equation.