Question

For what values of c, the roots of the equation $\left( {c - 2} \right){x^2} + 2\left( {c - 2} \right)x + 2 = 0$ are not real.
(A) $\left( {1,2} \right)$
(B) $\left( {2,3} \right)$
(C) $\left( {3,4} \right)$
(D) $\left( {2,4} \right)$

Answer 1

Hint: In the given question, we are required to solve for the values of c such that the equation $\left( {c - 2} \right){x^2} + 2\left( {c - 2} \right)x + 2 = 0$ has no real roots. We will first compare the given equation with the standard form of a quadratic equation $a{x^2} + bx + c = 0$ and then apply the condition over the discriminant over the discriminant of the quadratic equation.

Complete step by step answer:
In the given question, we are provided with the equation $\left( {c - 2} \right){x^2} + 2\left( {c - 2} \right)x + 2 = 0$.
Now, comparing the equation with standard form of a quadratic equation $a{x^2} + bx + c = 0$
Here,$a = \left( {c - 2} \right)$, \[b = 2\left( {c - 2} \right)\] and $c = 2$.
Now, we know that the roots of a quadratic equation are given by the quadratic formula $x = \dfrac{{( - b) \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. Now, if the roots are not real, the expression under the square root must be negative as then the roots will be complex.
The expression $\left( {{b^2} - 4ac} \right)$ present under the square root is called the discriminant of the quadratic equation.
So, for the roots to be non-real, the discriminant of the quadratic equation must be negative.
Hence, $D = \left( {{b^2} - 4ac} \right) < 0$
Substituting the values of a, b and c, we get,
$ \Rightarrow \left( {{{\left( {2\left( {c - 2} \right)} \right)}^2} - 4\left( {c - 2} \right)\left( 2 \right)} \right) < 0$
Opening the brackets,
$ \Rightarrow \left( {4{{\left( {c - 2} \right)}^2} - 8\left( {c - 2} \right)} \right) < 0$
Computing the square term,
$ \Rightarrow \left( {4\left( {{c^2} - 2\left( 2 \right)\left( c \right) + {2^2}} \right) - 8\left( {c - 2} \right)} \right) < 0$
$ \Rightarrow \left( {4\left( {{c^2} - 4c + 4} \right) - 8\left( {c - 2} \right)} \right) < 0$
Simplifying the inequality, we get,
$ \Rightarrow 4{c^2} - 16c + 16 - 8c + 16 < 0$
Adding up the like terms,
$ \Rightarrow 4{c^2} - 24c + 32 < 0$
Dividing both sides by $4$, we get,
$ \Rightarrow {c^2} - 6c + 8 < 0$
Factorising the quadratic expression,
$ \Rightarrow \left( {c - 2} \right)\left( {c - 4} \right) < 0$
Now, since the product of two terms is negative, one of the terms has to be negative and the other one has to be positive.
This gives rise to two cases.
Case $1$: $\left( {c - 2} \right) < 0$ and $\left( {c - 4} \right) > 0$
Simplifying the inequality using transposition,
$c < 2$ and $c > 4$
Now, both these inequalities don’t hold at the same time. So, the solution set for this case is null set.
Case $2$: $\left( {c - 2} \right) > 0$ and $\left( {c - 4} \right) < 0$
Simplifying the inequality using transposition,
$c > 2$ and $c < 4$
Now, both these inequalities hold at the same time for $c \in \left( {2,4} \right)$. So, the solution set for this case is $c \in \left( {2,4} \right)$.
Now, we have to take a union of solutions to both the cases. So, the final solution is $c \in \left( {2,4} \right)$.
Therefore, the values of c for which the equation $\left( {c - 2} \right){x^2} + 2\left( {c - 2} \right)x + 2 = 0$ has no real roots are $c \in \left( {2,4} \right)$.
So, the correct answer is “Option D”.

Note:
Quadratic equations are the polynomial equations with degree of the variable or unknown as 2. We should also know the expression of the discriminant of a quadratic equation so as to solve the question. We should take care of the calculations to be sure of the final answer. We should remember the expression for the discriminant of the quadratic equation.