Question

If the two roots of the equation, \[(a - 1)({x^4} + {x^2} + 1) + (a + 1){({x^2} + x + 1)^2} = 0\] are real and distinct, then set of all values of a is
A) \[(0,\dfrac{1}{2})\]
B) \[( - \dfrac{1}{2},0) \cup (0,\dfrac{1}{2})\]
C) \[( - \infty , - 2)\cup(2,\infty )\]
D) \[( - \dfrac{1}{2},0)\]

Answer 1

Hint: Firstly, break the term \[({x^4} + {x^2} + 1) = ({x^2} + x + 1)({x^2} - x + 1)\] and substitute it in the above equation and take common from both the parts of the provided equation. Then apply the condition for real and distinct root as for a quadratic equation \[D > 0\]. Hence, on solving the obtained inequality required solution will be obtained.

Complete step by step solution: The given equation is \[(a - 1)({x^4} + {x^2} + 1) + (a + 1){({x^2} + x + 1)^2} = 0\]
Now, split the term as \[({x^4} + {x^2} + 1) = ({x^2} + x + 1)({x^2} - x + 1)\] and then take common from both the parts as shown below,
\[(a - 1)({x^2} + x + 1)({x^2} - x + 1) + (a + 1){({x^2} + x + 1)^2} = 0\]
Take \[({x^2} + x + 1)\] common from both the parts as,
\[ \Rightarrow ({x^2} + x + 1)[(a - 1)({x^2} - x + 1) + (a + 1)({x^2} + x + 1)] = 0\]
Now, simplify the terms inside the bracket and convert it as it will form a quadratic equation, we get,
\[
   \Rightarrow ({x^2} + x + 1)[(a{x^2} - ax + a - {x^2} + x - 1) + (a{x^2} + ax + a + {x^2} + x + 1)] = 0 \\
   \Rightarrow ({x^2} + x + 1)[2a{x^2} + 2x + 2a] = 0 \\
 \]
Taking common and simplifying further the term can be represented as,
\[ \Rightarrow 2({x^2} + x + 1)[a{x^2} + x + a] = 0\]
As per the given condition, the equation has two roots as real and distinct.
So, calculating D for \[({x^2} + x + 1)\]
\[
  D = {b^2} - 4ac \\
   = {(1)^2} - 4(1)(1) \\
   = ( - 3) \\
 \]
It is cleared that \[{\text{D > 0}}\]definitely for \[(a{x^2} + x + a)\]as two roots are real and distinct for this equation.
 \[
  D = {b^2} - 4ac > 0 \\
   = {(1)^2} - 4(a)(a) > 0 \\
   = 1 - 4{a^2} > 0 \\
 \]
Now, solve the inequality in order to reach to final answer as,
\[
   \Rightarrow 4{a^2} - 1 < 0 \\
   \Rightarrow 4{a^2} < 1 \\
   \Rightarrow {a^2} < \dfrac{1}{4} \\
 \]
Hence form the above inequality we can determine that range of value of a is
\[ \Rightarrow - \dfrac{1}{2} < a < \dfrac{1}{2}\]
Now, in order to exist a quadratic equation it’s first term should be greater than zero as,
\[\therefore a \ne 0\]
Hence, we need to exclude zero from the obtained set of values of a and so our final answer will be,
\[( - \dfrac{1}{2},0)\cup(0,\dfrac{1}{2})\]

So, option (B) is our correct answer.

Note: If the discriminant is greater than zero, this means that the quadratic equation has two real, distinct (different) roots.
In algebra, a quadratic equation is an equation that can be rearranged in standard form as where x represents an unknown, and a, b, and c represent known numbers, where \[a \ne 0\]. Whereas, in coordinate geometry, a quadratic equation represents the parabola and root of a quadratic equation represents its intersection with the x-axis.