Question

The range of values of “a” for which all the roots of the \[\left( {a - 1} \right){\left( {1 + x + {x^2}} \right)^2} = \left( {a + 1} \right)\left( {1 + {x^2} + {x^4}} \right)\] imaginary is?
a) \[( - \infty , - 2]\]
b) \[(2, + \infty )\]
c) \[\left( { - 2,2} \right)\]
d) none of these

Answer 1

Hint: To solve this question, we first need to simplify the given equation into a simple form or standard form of a quadratic equation. Then, from that standard form we will find the discriminant of that quadratic equation. Then we will apply the condition for imaginary roots of a quadratic equation to find the range of values of “a”.

Complete answer:
We have the equation as:
\[\left( {a - 1} \right){\left( {1 + x + {x^2}} \right)^2} = \left( {a + 1} \right)\left( {1 + {x^2} + {x^4}} \right)\]
On rearranging the terms, we get;
\[ \Rightarrow \dfrac{{a - 1}}{{a + 1}} = \dfrac{{1 + {x^2} + {x^4}}}{{{{\left( {1 + x + {x^2}} \right)}^2}}}\]
Now we will try to express the numerator in the form of a perfect square. So, we can write;
\[ \Rightarrow \dfrac{{a - 1}}{{a + 1}} = \dfrac{{1 + {x^2} + {x^4} + {x^2} - {x^2}}}{{{{\left( {1 + x + {x^2}} \right)}^2}}}\]
Now we will group the terms. So, we have;
\[ \Rightarrow \dfrac{{a - 1}}{{a + 1}} = \dfrac{{\left( {1 + 2{x^2} + {x^4}} \right) - {x^2}}}{{{{\left( {1 + x + {x^2}} \right)}^2}}}\]
The numerator can be written as \[{\left( {a + b} \right)^2}\]. So, we have;
\[ \Rightarrow \dfrac{{a - 1}}{{a + 1}} = \dfrac{{{{\left( {1 + {x^2}} \right)}^2} - {x^2}}}{{{{\left( {1 + x + {x^2}} \right)}^2}}}\]
Now we will use the formula: \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]
So, we have;
\[ \Rightarrow \dfrac{{a - 1}}{{a + 1}} = \dfrac{{\left( {1 + x + {x^2}} \right)\left( {1 + {x^2} - x} \right)}}{{{{\left( {1 + x + {x^2}} \right)}^2}}}\]
On cancelling the terms, we get;
\[ \Rightarrow \dfrac{{a - 1}}{{a + 1}} = \dfrac{{\left( {1 + {x^2} - x} \right)}}{{\left( {1 + x + {x^2}} \right)}}\]
By applying the componendo-dividendo theorem we get;
\[ \Rightarrow \dfrac{{a - 1 - a - 1}}{{a - 1 + a + 1}} = \dfrac{{\left( {1 + {x^2} - x} \right) - \left( {1 + x + {x^2}} \right)}}{{\left( {1 + {x^2} - x} \right) + \left( {1 + x + {x^2}} \right)}}\]
On simplification we get;
\[ \Rightarrow \dfrac{{ - 2}}{{2a}} = \dfrac{{ - 2x}}{{2\left( {{x^2} + 1} \right)}}\]
On dividing by two we get;
\[ \Rightarrow \dfrac{1}{a} = \dfrac{x}{{\left( {{x^2} + 1} \right)}}\]
Now we will reciprocate. So, we have;
\[ \Rightarrow a = \dfrac{{\left( {{x^2} + 1} \right)}}{x}\]
On cross-multiplication we get;
\[ \Rightarrow ax = {x^2} + 1\]
On rearranging we get;
\[ \Rightarrow {x^2} + 1 - ax = 0\]
Now this is the standard form of the quadratic equation. The discriminant of this equation is:
\[D = {\left( { - a} \right)^2} - 4 \times 1 \times 1\]
On simplification;
\[ \Rightarrow D = {a^2} - 4\]
Now we know for imaginary roots, discriminant should be less than zero, so, we have;
\[ \Rightarrow {a^2} - 4 < 0\]
On factoring we get;
\[ \Rightarrow \left( {a + 2} \right)\left( {a - 2} \right) < 0\]
This gives
\[ \Rightarrow a \in \left( { - 2,2} \right)\]

Therefore, the correct option is (C)

Note: The quadratic equation we have has a positive coefficient of \[{x^2}\] so, the graph of this equation will be a parabola with upward facing or upward concavity. Another thing to note is that the imaginary roots always occur in pairs.