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# If the functions $f\left( x \right)=a{{x}^{2}}+bx+c$ and $g\left( x \right)=-a{{x}^{2}}+bx+c$ where $ac\ne 0$ then $f\left( x \right).g\left( x \right)=0$ has(A) At least two real roots(B) At least three real roots(C) At least one real roots(D) At least two imaginary roots

Last updated date: 17th Sep 2024
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Hint: We start solving this problem by first finding the discriminant of both the functions $f\left( x \right)$ and $g\left( x \right)$ using the formula, discriminant of equation $a{{x}^{2}}+bx+c=0$ is ${{b}^{2}}-4ac$. Then we assume the cases when ac is greater than zero and find the nature of the discriminant of both functions and then we assume that ac is less than zero and find the nature of discriminant again. From the obtained natures of discriminants we can find the nature of roots of the function $f\left( x \right).g\left( x \right)=0$.

For any quadratic equation $a{{x}^{2}}+bx+c=0$, nature of its roots can be said by the discriminant of the equation, that is ${{b}^{2}}-4ac$.
Roots are real and distinct if ${{b}^{2}}-4ac>0$.
Roots are equal if ${{b}^{2}}-4ac=0$.
Roots are imaginary if ${{b}^{2}}-4ac<0$.

We are given that $f\left( x \right).g\left( x \right)=0$. The possible roots for it are roots of $f\left( x \right)=0$ and roots of $g\left( x \right)=0$.
Roots of $f\left( x \right)=0$ are real if its discriminant is greater than zero.
Discriminant of $f\left( x \right)=a{{x}^{2}}+bx+c$ is ${{b}^{2}}-4ac$.
It has real roots if ${{b}^{2}}-4ac>0$
Similarly roots of $g\left( x \right)=0$ are real if its discriminant is greater than zero.
Discriminant of $g\left( x \right)=-a{{x}^{2}}+bx+c$ is ${{b}^{2}}-4\left( -a \right)c={{b}^{2}}+4ac$.
It has real roots if ${{b}^{2}}+4ac>0$
We are given that $ac\ne 0$, so we have two possibilities $ac>0$ or $ac<0$.
If $ac>0$, then
Discriminant of $g\left( x \right)=-a{{x}^{2}}+bx+c$, ${{b}^{2}}+4ac$ is greater than zero, because
\begin{align} & \Rightarrow ac>0 \\ & \Rightarrow 4ac>0 \\ & \Rightarrow {{b}^{2}}+4ac>{{b}^{2}}>0 \\ & \Rightarrow {{b}^{2}}+4ac>0 \\ \end{align}
It has two real roots.
Discriminant of $f\left( x \right)=a{{x}^{2}}+bx+c$, ${{b}^{2}}-4ac$ it cannot be estimated. So, it can have either real or imaginary roots.
If $ac<0$, then
Discriminant of $f\left( x \right)=a{{x}^{2}}+bx+c$, ${{b}^{2}}-4ac$ is greater than zero, because
\begin{align} & \Rightarrow ac<0 \\ & \Rightarrow 4ac<0 \\ & \Rightarrow -4ac>0 \\ & \Rightarrow {{b}^{2}}-4ac>{{b}^{2}}>0 \\ & \Rightarrow {{b}^{2}}-4ac>0 \\ \end{align}
It has two real roots.
Discriminant of $g\left( x \right)=-a{{x}^{2}}+bx+c$, ${{b}^{2}}+4ac$ cannot be estimated. So, it can have either real or imaginary roots.
In both the cases one of the functions has real roots and the nature of the other is unknown.
So, we can say that the function $f\left( x \right).g\left( x \right)=0$ has two real roots and we don’t know the nature of the other two roots.
So, we can say that the function $f\left( x \right).g\left( x \right)=0$ has at least two real roots.
Note:The main mistake one makes while solving this problem is one might not consider the other equation after finding the nature of roots of one function and might take the answer as, that the function $f\left( x \right).g\left( x \right)=0$ has exactly two roots.