Question

let a and b be two positive real numbers such that a√a+b√b=32and a√b+b√a=31.what is the value of 5(a+b)/7?​

Answer 1

Answer:

9

Step-by-step explanation:

Given :

a√a + b√b = 32,

a√b + b√a = 31.

Solution :

a√a + b√b = 32  ...(1)

a√b + b√a = 31  ...(2)

let a = x² & b = y².

Then,

(1) can be re-written as,

⇒ (x)³ + (y)³ = 32 ....(3)

(2) can be re-written as,

⇒ x²y + xy² = 31 ...(4)

by adding  3×(4) + (3),

⇒ x³ + y³ + 3x²y + 3xy² = 32 + 31×3 = 125

⇒ (x + y)³ = 125

⇒ x + y = 5   ( ∵ -5 is not possible, as (-5)³ = -125 )

from (4)

⇒ x²y + xy² = 31,

⇒ x²y + xy² = xy(x + y) = 31

⇒ 5xy = 31

⇒ $xy=\frac{31}{5}$

⇒ a + b = x² + y² = (x + y)² - 2xy

= $5^2-2(\frac{31}{5})=25-\frac{62}{5}=\frac{125-62}{5}=\frac{63}{5}$

so,

$\frac{5(a+b)}{7} = \frac{5(\frac{63}{5})}{7}= \frac{63}{7} = 9$

Is this question from PRMO 2017 (I remember like solving this Q.)

Answer 2

Answer:

9

Step-by-step explanation:

Given :

a√a + b√b = 32,

a√b + b√a = 31.

Solution :

a√a + b√b = 32 ...(1)

a√b + b√a = 31 ...(2)

let a = x² & b = y².

Then,

(1) can be re-written as,

⇒ (x)³ + (y)³ = 32 ....(3)

(2) can be re-written as,

⇒ x²y + xy² = 31 ...(4)

by adding 3×(4) + (3),

⇒ x³ + y³ + 3x²y + 3xy² = 32 + 31×3 = 125

⇒ (x + y)³ = 125

⇒ x + y = 5 ( ∵ -5 is not possible, as (-5)³ = -125 )

from (4)

⇒ x²y + xy² = 31,

⇒ x²y + xy² = xy(x + y) = 31

⇒ 5xy = 31

⇒ xy=\frac{31}{5}xy=

5

31

⇒ a + b = x² + y² = (x + y)² - 2xy

= 5^2-2(\frac{31}{5})=25-\frac{62}{5}=\frac{125-62}{5}=\frac{63}{5}5

2

−2(

5

31

)=25−

5

62

=

5

125−62

=

5

63

so,

\frac{5(a+b)}{7} = \frac{5(\frac{63}{5})}{7}= \frac{63}{7} = 9

7

5(a+b)

=

7

5(

5

63

)

=

7

63

=9

Is this question from PRMO 2017 (I remember like solving this Q.)