Question

If zeroes of the polynomial x² + (p + 1)x + q are 2 and -

3, then find value of p​

Answer 1

Step-by-step explanation:

Given:-

Zeros of the polynomial $x^2+(p+1)x+q$are 2 and -3

To find:-

Value of P

Solution:-

Let the zeros are =$\alpha,\beta$

• Now

$\sf \alpha+\beta=\dfrac {-b}{a}$

• Substitute the values

${:}\rightarrowtail$$\sf 2+(-3)=\dfrac {(p+1)}{1}$

${:}\rightarrowtail$$\sf 2-3=p+1$

${:}\rightarrowtail$$\sf p+1=-1$

${:}\rightarrowtail$$\sf p=-1+1$

${:}\rightarrowtail$$\sf p=0$

$\therefore\underline{\underline{\sf Value\:of\:p\:is\:0.}}$

Answer 2

Answer:

Given equation

x

2

+px+q

Also, given p,q are the roots of the equation.

p+q=−p

2p=−q .....(1)

And pq=q

q(p−1)=0

⇒q=0 or p=1

So, by (1), q=0⇒p=0

Hence, the values of p are 0,1.

hope it may help you :)