Question

Find two positive numbers whose difference is 12 and whose am exceeds the gm by 2

Answer 1

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We will let the two numbers be a and b as the positive numbers:

a - b = 12 (this will be equation 1)

The A.M will exceed G.M by around 2

a + b/2 = √ ab + 2   =   a + b = 2 √ab + 4

a + b - 2 √ab = 4    = (√a -√b)² = 4

√a - √b = 2 (this will be equation 2)

From the equation 1) (√a - √b) (√a + √b) = 12  

=  2 (√a + √b)  = 12  

= √a + √b = 6 (this will be equation 3)

now we will add 2 and 3

2 √a = 8

√a  = 4

a = 16

now we will subtract 2 from 3

we will get 2√b = 4

= √b = 2

b = 4 is the final answer

if there is any confusion please leave a comment below.

Answer 2

AnswEr:

Let the two numbers be a and b such that a > b . It is given that

$\qquad\sf{a-b=12-\:-\:-\:(i)}$

It is also given that

$\qquad\rm{AM-GM=2}$

$\rightarrow \sf \: \frac{a + b}{2} - \sqrt{ab} = 2 \\ \\ \rightarrow \sf \: a + b - 2 \sqrt{ab} = 4 \\ \\ \rightarrow \sf \: ( \sqrt{a} - \sqrt{b} ) {}^{2} = 4 \\ \\ \rightarrow \sf \: \sqrt{a} - \sqrt{b} = 2 - - (ii)\\ \\ \rm \: now, \qquad \: \: \: \: \sf \: a - b = 12 \\ \\ \implies \sf \: ( \sqrt{a} + \sqrt{b} )( \sqrt{a} - \sqrt{b} ) = 12 \\ \\ \implies \sf \: ( \sqrt{a} + \sqrt{b} ) \times (2) = 12 - - - (iii) \\ \\ \implies \sf \: \sqrt{a} + \sqrt{b} = 6 \:$

Solving (ii) and (iii) , we get a = 16, b = 4.

Hence, the required numbers are 16 and 4. ,