Question

If the roots of the equation \[b{{x}^{2}}+cx+a=0\] be imaginary, then for all real values of x, the expression \[3{{b}^{2}}{{x}^{2}}+6bcx+2{{c}^{2}}\] is
(a) greater than 4ab
(b) less than 4ab
(c) greater than -4ab
(d) less than -4ab

Answer 1

Hint: This question belongs to the topic of quadratic equations and polynomials. If we want to solve this question, we are given a quadratic equation, and a second-degree polynomial of a single variable. First of all, we will use the quadratic equation, and form a condition with the given data about imaginary roots of the equation. Then we will use the condition to find the value of the second-degree polynomial of a single variable.

Complete step-by-step answer:Now, we will solve the complete question.
We will start the solution by taking the quadratic equation.
The quadratic equation given in the question is,
\[b{{x}^{2}}+cx+a=0\]
According to the question, the roots of the quadratic equation are imaginary.
So,
\[\begin{align}
  &\Rightarrow {{c}^{2}}-4ab<0 \\
 &\Rightarrow -{{c}^{2}}>-4ab \\
\end{align}\]
We will note this equation as equation (i)
Now, we take the polynomial
The polynomial given in question is,
\[3{{b}^{2}}{{x}^{2}}+6bcx+2{{c}^{2}}\]
Let us assume that the value of a polynomial is y, where y is a variable.
So,
\[3{{b}^{2}}{{x}^{2}}+6bcx+2{{c}^{2}}=y\]
This equation can also be written as,
\[3{{b}^{2}}{{x}^{2}}+6bcx+2{{c}^{2}}-y=0\]
According to the question, we need to find the value of a polynomial when the value of x is real.
So,
\[{{(6bc)}^{2}}-4\times (3{{b}^{2}})(2{{c}^{2}}-y)\ge 0\]
Further solving it, we will get,
\[\begin{align}
  &\Rightarrow {{(6bc)}^{2}}-4\times (3{{b}^{2}})(2{{c}^{2}}-y)\ge 0 \\
 &\Rightarrow 36{{b}^{2}}{{c}^{2}}-12{{b}^{2}}(2{{c}^{2}}-y)\ge 0 \\
 &\Rightarrow 3{{c}^{2}}-2{{c}^{2}}+y\ge 0 \\
 &\Rightarrow {{c}^{2}}+y\ge 0 \\
 &\Rightarrow y\ge -{{c}^{2}} \\
\end{align}\]
We will note this equation as equation (ii)
Now, if we compare equation (i) with equation (ii), we will get,
\[\begin{align}
  &\Rightarrow y>-4ab \\
 &\Rightarrow 3{{b}^{2}}{{x}^{2}}+6bcx+2{{c}^{2}}>-4ab \\
\end{align}\]

So, the correct answer is “Option C”.

Note:The first thing that will confuse students is the opposite conditions of x, in the quadratic equation and polynomial. In the quadratic equation, it is said that the values of x are imaginary, whereas in the polynomial we need to find the value when the values of x are real. So, for this, we just need to focus on the condition and not on words.