Question

If $(p+q)x^4+p{x}^2+6x+1$ is a quadratic polynomial, then the value of $p,q$ can be:

• A. $p=0,q=0$
• B. $p=3,q=-2$
• C. $p=1,q=-1$
• D. $p=-2,q=2$

Answer 1

Answer: (C), (D)

$p=-2,q=2$

Given that $(p+q)x^4+p{x}^2+6x+1$ is a quadratic polynomial.
For this to be a quadratic polynomial, the highest power of the varible must be $2$.
$\Rightarrow$ the coefficients of the $x^4$ term must be zero.
$\Rightarrow p+q=0$
Also since $p$ is the coefficient of $x^2$ term, it cannot be zero.
$\Rightarrow p\ne 0$
From the options,
$\left(a\right)p=0,q=0$ is not possible as $p\ne 0$.

$\left(b\right)p=1,q=-1$ is possible as $p+q=0,p\ne 0$

$\left(c\right)p=-2,q=2$ is possible as $p+q=0,p\ne 0$

$\left(d\right)p=3,q=-2$ is not possible as $p+q\ne 0$