Question

If one zero of quadratic polynomial x square +(p+1)x-6 is 2,then what is the value of P? [sol:p=0]​

Answer 1

SOLUTION

Let the given polynomial be

$\sf \: p(x) = {x}^{2} + (p + 1)x - 6$

One of the zeros of the polynomial is 2

NoTE

• Sum of Zeros : - x coefficient/x² coefficient

• Product of Zeros : constant term /x² coefficient

Let the other zero be a

Here,

$\tt \: 2a = - 6 \\ \\ \longrightarrow \: \boxed{\boxed{ \tt a = - 3}}$

The zeros of the polynomial are 2 and - 3

Also,

$\tt \: \longrightarrow 2 + ( - 3) = - \dfrac{(p + 1)}{1} \\ \\ \longrightarrow \: \tt \: -1 = - (p + 1) \\ \\ \large{\longrightarrow\: \boxed{ \boxed{ \tt \: p = 0}}}$

Answer 2

Given equation,

$x^{2} +(p+1)x - 6 = 0$

One of the zeroes of the polynomial is 2

Substituting, x = 2, in the first equation, we get,

$(2)^{2} +(p+1)2 - 6 = 0$

$4 + 3p + 2 - 6 = 0$

$3p + 6 - 6 = 0$

$3p = 0$

$\underline{\underline{\bold{p=0}}}$

$\boxed{\boxed{\bold{Therefore, \ the \ value \ of \ p \ is \ 0.}}}$