Question

Let S$= \\{ x \in R:x \geqslant 0\,and\,2\left| {\sqrt x - 3} \right| + \sqrt x (\sqrt x - 6) + 6 = 0\\}$. Then S
A) Contains exactly two elements.
B) Contains exactly four elements
C) Is an empty set.
D) Contains exactly one element.

Answer 1

This question can be solved by substitution. Substitute $\sqrt x = y$ or$\sqrt x - 3 = y$or whichever you feel like and write the given equation in terms of $y$. Further, solve the equation and get the values of $y$ from which you can get the values of $x$

Complete step-by-step solution:
Here in this problem, we are given with definition of a set S which contains non-negative real numbers ‘x’ which follows the equation $2\left| {\sqrt x - 3} \right| + \sqrt x (\sqrt x - 6) + 6 = 0$ . Using this information we need to correct the option among the given four choices.
First, we will substitute $\sqrt x - 3 = y$
But why? This is because solving the equation with$\sqrt x$will be a bit difficult. So we replace it by $y$ and will solve it in terms of $y$
$\Rightarrow \sqrt x - 3 = y$
Using the value of $\sqrt x$ in the given equation, we get:
$\Rightarrow 2\left| y \right| + (y + 3)(y - 3) + 6 = 0$
Using identity $(a + b)(a - b) = {a^2} - {b^2}$ in the above equation, we can solve it further:
$\Rightarrow 2\left| y \right| + {y^2} - 9 + 6 = 0 \Rightarrow {y^2} + 2\left| y \right| - 3 = 0$
As ${y^2}$is always positive, hence we can write ${y^2} = {\left| y \right|^2}$
$\Rightarrow {\left| y \right|^2} + 2\left| y \right| - 3 = 0$
This becomes a quadratic equation; with the product as minus three and sum two we get:
$\Rightarrow {\left| y \right|^2} + 3\left| y \right| - \left| y \right| - 3 = 0 \Rightarrow \left| y \right|\left( {\left| y \right| + 3} \right) - \left( {\left| y \right| + 3} \right) = 0$
After taking common, we can represent the above equation as:
$\left( {\left| y \right| - 1} \right)\left( {\left| y \right| + 3} \right) = 0$
Hence we get:
$\Rightarrow \left| y \right| = 1{\text{ or }}\left| y \right| = - 3$
As $\left| y \right|$ cannot be negative therefore, $\left| y \right| = - 3$is rejected
Therefore, $\left| y \right| = 1 \Rightarrow y = \pm 1$
Now we have got the value of $y$
But we need to find the values of $x$ and so we will replace $y$ by $\sqrt x - 3$
$\Rightarrow \sqrt x - 3 = 1\,\,\,and\,\,\,\sqrt x - 3 = - 1 \Rightarrow \sqrt x = 4\,\,and\,\,\sqrt x = 2$
By squaring both sides of the above equation, we get
$\Rightarrow x = 16\,\,,\,\,x = 4$
Therefore, the set S contains only two elements, i.e. $16{\text{ and 4}}$ .

Hence, option A is correct

Note: In such questions, substitution is very important, and choosing the correct substitution is more important. Although in the question we solved, it hardly matters whether we choose $\sqrt x$ or $\sqrt x - 3$ or $\sqrt x - 6$ for substitution. This means you need to substitute the correct variable i.e. variable and ignore the coefficients. An alternative approach can be taken to solve the quadratic equation ${\left| y \right|^2} + 2\left| y \right| - 3 = 0$ by using the quadratic formula, i.e. $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ .