Question

The set $A = \left\{ {x:x \in {x^2} = 16 and 2x = 6} \right\}$ equals
$\eqalign{
  & 1)\emptyset \cr
  & 2)\left\{ {14,3,4} \right\} \cr
  & 3)\left\{ 3 \right\} \cr
  & 4)\left\{ 4 \right\} \cr} $

Answer 1

Hint: In the given question, a set $A$ is defined. It is given that $x$ belongs to ${x^2} = 16$ and $2x = 6$ . So, we need to find the values of $x$ in each of these cases. So, these values will form the set. If no values are appropriate, then it will be a null set. A null set is a set which does not have any potential values for the given condition. It is also called an empty set.

Complete step-by-step answer:
Let us consider the given set, which is
$A = \left\{ {x:x \in {x^2} = 16and2x = 6} \right\}$
Now, we need to find out the elements inside the set. Since two equations are given, we can find out the elements.
The first one is,
${x^2} = 16$
By removing the squares on both the sides, we have
$x = \pm 4$
This is the value of the first element.
Next, we find out the value of the second element which is given by the equation,
$2x = 6$
Rearranging the equation, we get
$x = 3$
There exist no elements in between these two elements.
Since no value of $x$is satisfied, the given set will be a null set.
That is, $A = \left\{ \emptyset \right\}$
Therefore, the final answer is $\emptyset $.
Hence, option (1) gives us the correct answer.
So, the correct answer is “Option 1”.

Note: All sets need not have values, some sets can be empty. It is given that all values $x$ belong to the set. Therefore, we need to find the values by solving those equations. While removing the square root, do not forget that the number gets both positive and the negative sign.