Questions & Answers

Question

Answers

Has at least

(a) Four real roots

(b) Two real roots

(c) Four imaginary roots

(d) None of these

Answer
Verified

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.

Sometimes, a student makes a mistake by using the discriminant formula that is ${{B}^{2}}-4AC$. So, Students have to take care of it.