Question

If $p,q,r,s\in R$ , then equation $\left(x^2+px+3q\right)\left(-x^2+rx+q\right)\left(-x^2+sx-2q\right)=0$ has:
A. $6$ real roots
B.At least two real roots
C. $2$ real and $4$ imaginary roots
D. $4$ real and $2$ imaginary roots

Answer 1

Hint: Check the nature of the roots of the given equation by taking each quadratic polynomial equal to zero and check the nature of the roots of each quadratic equation.

Complete step-by-step answer:
The given equation
$\left(x^2+px+3q\right)\left(-x^2+rx+q\right)\left(-x^2+sx-2q\right)=0$ is true if $x^2+px+3q=0$ or $-x^2+rx+q=0$ or $-x^2+sx-2q=0$ for some value of $x$ .
Now check the nature of the roots of the first quadratic equation by checking the value of $b^2-4ac$ .
The value of $b^2-4ac$ for first quadratic polynomial is:
 $b^2-4ac=p^2-12q$ .
As the numbers $p,q\in R$ . So, the value of $p^2-12q$ can be less than or greater than or equal to zero for different values of $p$ and $q$ .
So, the roots of the quadratic equation $x^2+px+3q=0$ can be distinct real, same real or imaginary numbers.
Now check the nature of the roots of the second quadratic equation by checking the value of $b^2-4ac$ .
The value of $b^2-4ac$ for second quadratic polynomial is:
 $$\begin{gathered} b^2-4ac=r^2-4\left(-1\right)q \\ =r^2+4q-4rq+4rq \\ ={\left(r-2q\right)}^2+4rq \end{gathered}$$ .
As the numbers $r,q\in R$ . So, the value of ${\left(r-2q\right)}^2+4rq$ can be greater than zero for different values of $r$ and $q$ .
So, the roots of the quadratic equation $-x^2+rx+q=0$ can be distinct real numbers.
Now check the nature of the roots of the third quadratic equation by checking the value of $b^2-4ac$ .
The value of $b^2-4ac$ for third quadratic polynomial is:
 $$\begin{gathered} b^2-4ac=s^2-4\left(-1\right)\left(-2q\right) \\ =s^2-8q \end{gathered}$$ .
As the numbers $s,q\in R$ . So, the value of $s^2-8q$ can be less than or greater than or equal to zero for different values of $s$ and $q$ .
So, the roots of the quadratic equation $-x^2+sx-2q=0$ can be distinct real, same real or imaginary numbers.
So, the equation $\left(x^2+px+3q\right)\left(-x^2+rx+q\right)\left(-x^2+sx-2q\right)=0$ has at least $2$ real roots.
So, the correct answer is “Option B”.

Note: The equation $\left(x^2+px+3q\right)\left(-x^2+rx+q\right)\left(-x^2+sx-2q\right)=0$ is true if and only if any one of the factor of the equation is equal to zero. Also the nature of the roots of the given equation is the same as the nature of the roots of the quadratic equations.