Question

A.M between two numbers is 6 . the sum of their squars is 90. find the numbers​

Answer 1

Answer:

The numbers are 9 and 3

Step-by-step explanation:

Let the two numbers be "a" and "b"

Given:

AM of a and b = 6

=> (a + b)/2 = 6

=> (a + b) = 12 ....(i)

Sum of squares of "a" and "b" = 90

=> a² + b² = 90 ..(ii)

From (i),

(a + b)² = 144

=> a² + 2ab + b² = 144

Using (ii),

=> 90 + 2ab = 144

=> 2ab = 144 - 90

=> 2ab = 54

=> ab = 27 ............(iii)

From (iii),

b = 27/a

Using this in (i), we get:

a + 27/a = 12

a² + 27 = 12a

a² - 12a + 27 = 0

(a - 9)(a + 3) = 0

a = 9 or a = -3

As numbers are assumed to be positive, a = -3 is inadmissible.

=> a = 9

=> b = 27/a = 27/9 = 3

The numbers are 9 and 3

Verify:

AM of 9 and 3 is 6

Sum of squares of 9 and 3 = 81 + 9 = 90

Answer 2

The two numbers are 9 and 3.

• AM of two numbers is given as 6.
• Let the two numbers be a and b.

$\frac{a+b}{2}$ = 6

a+b =12

• Sum of their square is given as 90.

$a^{2}  + b^{2}$ = 90

• As we have an equation in two variable of degree 1, we need one more equation in two variable with degree 1 to solve the equation and find the solution.

• For finding the other equation, we use the identity of $(a+b)^{2} = a^{2} + 2ab + b^{2}$
• Substituting the known values,

$12^{2} = 90 + 2ab$

2ab = 54

ab = 27

• Now we use other identity $(a-b)^{2} = a^{2} - 2ab + b^{2}$

$(a-b)^{2} = 90 - 54$

$(a-b)^{2} = 36$

we get , (a-b) =6 or (a-b)= -6

• first we use a-b =6 to find the solution
• We have two equations a+b =12 and a-b =6
• On solving both the equations we get a=9 and b=3
• Therefore the two numbers are 9 and 3.
• (By using the other equation a-b= -6, we get a=3 and b=9, Therefore the solution remains the same!)