Question

What should come in place of both x in the equation $ \dfrac{x}{{\sqrt {128} }} = \dfrac{{\sqrt {162} }}{x}? $
A) \[12\]
B) \[14\]
C) \[144\]
D) \[196\]

Answer 1

Hint: To solve this problem, i.e., to find the value of x in the given equation. We will first take variables in one side and constants in one side. At L.H.S. we will get a square of x, and at the R.H.S. we will get the square root of numbers. Since, the given constants are the square root of numbers. After opening the square root, we will get the value of \[{x^2}.\] Then, again taking the square root on both the sides, we will get the value of x.

Complete step-by-step answer:
We need to find what should come in the place of both x in the equation given to us, i.e.., $ \dfrac{x}{{\sqrt {128} }} = \dfrac{{\sqrt {162} }}{x}. $
So the given equation is, $ \dfrac{x}{{\sqrt {128} }} = \dfrac{{\sqrt {162} }}{x}. $
On taking, variables at one side and constants at another side, we get
 $\Rightarrow {x^2} = \sqrt {(128) \times (162)} $
Now, on taking the multiplication of those number which are square of some numbers, (so that we can remove the square root), we get
 $\Rightarrow {x^2} = \sqrt {(64 \times 2) \times (18 \times 9)} $
On arranging the numbers, so that squares of numbers are produced, we get
 $
\Rightarrow {x^2} = \sqrt {64 \times 36 \times 9} \\
\Rightarrow {x^2} = \sqrt {{8^2} \times {6^2} \times {3^2}} \\
  $
Now on removing the square root on the R.H.S., we get
 $
\Rightarrow {x^2} = 8 \times 6 \times 3 \\
{x^2} = 144 \\
  $
On taking, square root on both sides, we get
 $ \Rightarrow x = 12...........(\because \sqrt {144} = 12) $
Thus, option (A) \[12,\] is correct.

So, the correct answer is “Option A”.

Note: Here, students might get confused that they need to find the value of \[{x^2},\] and they might choose the option \[144.\] But no, we are supposed to find the value of x, so the correct option will be option A, i.e., \[12,\] not option C, i.e., \[144.\]