Question
Answers

If $P\left( x \right)=a{{x}^{2}}+bx+c$ and $Q\left( x \right)=-a{{x}^{2}}+dx+c$ , where$ac\ne 0$ ,then $P\left( x \right)Q\left( x \right)=0$
Has at least
 (a) Four real roots
(b) Two real roots
(c) Four imaginary roots
(d) None of these

Answer Verified Verified
Hint: To solve the above question, we have to know the concept of quadratic equation. Let we consider a quadratic equation as $A{{x}^{2}}+Bx+C=0$ where the discriminant is ${{B}^{2}}-4AC$ .We have to know if in any quadratic equation, coefficient of ${{x}^{2}}$ and constant term are of opposite sign$\left( AC<0 \right)$ , then that equation will definitely have real roots.

Complete step by step answer:
Now given that,
$P\left( x \right)=a{{x}^{2}}+bx+c$
And $q(x)=-a{{x}^{2}}+dx+c=0$
Now we will consider the equation, $A{{x}^{2}}+Bx+C=0$ where if we find the value of $x$
Then we use the formula of $\dfrac{-B\pm \sqrt{{{B}^{2}}-4AC}}{2A}$ .
Here the discriminant is ${{B}^{2}}-4AC$ .
We can see that in the above equation $A$ and $C$ are of the opposite sign that is $AC$ is negative. So, we can see that ${{B}^{2}}-4AC$ will be positive and hence we also see that the equation will have real roots.
So, we have to know if in any quadratic equation, coefficient of ${{x}^{2}}$ and constant term are of opposite sign$\left( AC<0 \right)$ , then that Equation will definitely have real roots.
Now we take the $P\left( x \right)=a{{x}^{2}}+bx+c$
Where we can see that product of coefficient of ${{x}^{2}}$ and constant term $=ac$
And for $q(x)=-a{{x}^{2}}+dx+c=0$
Where we can see that product of coefficient of ${{x}^{2}}$ and constant term $=-ac$
So, we can see that if $ac$ is positive, then $-ac$ will be negative and vice versa.
And for this case one of the equations will definitely have real roots.
That is $\left( a{{x}^{2}}+bx+c \right)\left( -a{{x}^{2}}+dx+c \right)=0$ will have at least two real roots.

So, the correct answer is “Option B”.

Note: Here student must take care of the concept of discriminant and also take care of if coefficient of ${{x}^{2}}$ and constant term are of opposite sign$\left( AC<0 \right)$ , then that Equation will definitely have real roots.
Sometimes, a student makes a mistake by using the discriminant formula that is ${{B}^{2}}-4AC$. So, Students have to take care of it.