Question:

Two numbers $x$ and $y$ have arithmetic mean $9$ and geometric mean $4$. Then $x$ and $y$ are the roots of

Updated On: Jun 7, 2024
  • $x^2 - 18x - 16 = 0$
  • $x^2 - 18x + 16 = 0$
  • $x^2 + 18x - 16 = 0$
  • $x^2 + 18x+ 16= 0$
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The Correct Option is B

Solution and Explanation

We have,
$AM$ of $x, y=9$ and $GM$ of $x, y=4$
$\therefore \frac{x+y}{2}=9$ and $\sqrt{x y}=4$
$\Rightarrow x+y=18$ and $x y=16$
$\Rightarrow y=18-x$ and $x y=16$
$\therefore x(18-x)=16$
$\Rightarrow 18 x-x^{2}=16$
$\Rightarrow x^{2}-18 x+16=0$
$\therefore x$ and $y$ are the roots of the equation
$x^{2}-18 x+16=0$
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Concepts Used:

Relationship Between A.M. and G.M.

The relation between AM, GM, and HM can be acknowledged from the statement that the value of AM is greater than the value of GM and HM. For the same provided set of data points, the arithmetic mean is greater than the geometric mean, and the geometric mean is further greater than the harmonic mean. This relation between AM, GM, and HM can be conferred as the following expression.

\(AM > GM > HM\)

The arithmetic mean is also called the average of the stated numbers, and for two numbers a, and b, the arithmetic mean is equivalent to the sum of the two numbers, divided by 2.

\(AM = \frac{(a+b)}{2}\)

further, if there are 'n' numbers of data, then their geometric mean is equivalent to the nth root of the product of the ‘n’ numbers.

\(GM = \sqrt{ab}\)