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Question: The range of values of “a” for which all the roots of the \[\left( {a - 1} \right){\left( {1 + x + {...

The range of values of “a” for which all the roots of the (a−1)(1+x+x2)2=(a+1)(1+x2+x4)\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) (−∞,−2]( - \infty , - 2]
b) (2,+∞)(2, + \infty )
c) (−2,2)\left( { - 2,2} \right)
d) none of these

Explanation

Solution

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:
(a−1)(1+x+x2)2=(a+1)(1+x2+x4)\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;
⇒a−1a+1=1+x2+x4(1+x+x2)2\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;
⇒a−1a+1=1+x2+x4+x2−x2(1+x+x2)2\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;
⇒a−1a+1=(1+2x2+x4)−x2(1+x+x2)2\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 (a+b)2{\left( {a + b} \right)^2}. So, we have;
⇒a−1a+1=(1+x2)2−x2(1+x+x2)2\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: a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)
So, we have;
⇒a−1a+1=(1+x+x2)(1+x2−x)(1+x+x2)2\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;
⇒a−1a+1=(1+x2−x)(1+x+x2)\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;
⇒a−1−a−1a−1+a+1=(1+x2−x)−(1+x+x2)(1+x2−x)+(1+x+x2)\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;
⇒−22a=−2x2(x2+1)\Rightarrow \dfrac{{ - 2}}{{2a}} = \dfrac{{ - 2x}}{{2\left( {{x^2} + 1} \right)}}
On dividing by two we get;
⇒1a=x(x2+1)\Rightarrow \dfrac{1}{a} = \dfrac{x}{{\left( {{x^2} + 1} \right)}}
Now we will reciprocate. So, we have;
⇒a=(x2+1)x\Rightarrow a = \dfrac{{\left( {{x^2} + 1} \right)}}{x}
On cross-multiplication we get;
⇒ax=x2+1\Rightarrow ax = {x^2} + 1
On rearranging we get;
⇒x2+1−ax=0\Rightarrow {x^2} + 1 - ax = 0
Now this is the standard form of the quadratic equation. The discriminant of this equation is:
D=(−a)2−4×1×1D = {\left( { - a} \right)^2} - 4 \times 1 \times 1
On simplification;
⇒D=a2−4\Rightarrow D = {a^2} - 4
Now we know for imaginary roots, discriminant should be less than zero, so, we have;
⇒a2−4<0\Rightarrow {a^2} - 4 < 0
On factoring we get;
⇒(a+2)(a−2)<0\Rightarrow \left( {a + 2} \right)\left( {a - 2} \right) < 0
This gives
⇒a∈(−2,2)\Rightarrow a \in \left( { - 2,2} \right)

Therefore, the correct option is (C)

Note: The quadratic equation we have has a positive coefficient of x2{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.