Question

a√a + b√b =183 & a√b + b√a = 183 find the value of 9(a+b)/5

Answer 1

Given :
a√a + b√b =183 & a√b + b√a = 183

To Find :9(a+b)/5

Solution:

Let us assume √a = x and √b =y

⇒a = x²  and b = y²

Now let us consider :

a√a + b√b =183 

⇒x².x + y².y =183

⇒x³ + y³ =183 -----------equation(1)

Similarly a√b + b√a = 183 we can write as 

⇒x².y+y².x=183

⇒xy(x+y)=183 -----------equation(2)

(x+y)³ = x³ + y³ + 3xy(x+y)

⇒(x+y)³ -3xy(x+y) = x³ + y³

⇒(x+y)³ -3*xy*183/xy = 183

⇒(x+y)³- 549 =183

⇒(x+y)³ =732

⇒x+y= 3√732 ≈9.01≈9.0

∴x+ y =9

(x³ + y ³ )=(x+y) (x²+y²-xy)
(x³ + y ³ )=(x+y)(x²+y²) -xy(x+y)

⇒183=9(x²+y²) -183

9(x²+y²)-183=183

9(x²+y²) =366----------equation 3

Required value 9/5 (a+b)

(9/5)[x² +y²]

⇒(1/5)9[x² +y²]

From equation 3

⇒366/5 =73.2

∴the value of 9/5[a+b] = 73.2

Answer 2

Solution:

Given :
a√a + b√b =183 ………(1)

a√b + b√a = 183……….(2)

Let √a = x & √b =y


a = x² & b = y²

Put the value of a & √a , b, √b eq 1


a√a + b√b =183



(x² × x )+ (y² × y) =183

x³ + y³ =183 ………………..(3)


Put the value of a & √a , b, √b eq 2

a√b + b√a = 183


(x² × y)+(y² × x) =183

x²y + xy²= 183

xy(x+y)=183 ………………….(4)

[(x+y)³ = x³ + y³ + 3xy(x+y)]

(x+y)³ -3xy(x+y) = x³ + y³

(x+y)³ -3(183)= 183

[ From eq 3 & 4]

(x+y)³- 549 =183


(x+y)³ =183 + 549

(x+y)³ =732

x+y= ³√732 ≈9.01

x+ y =9(approximately)


xy(x+y) =183. (From eq 4)

xy (9)= 183 (x+y=9)

xy = 183/9……………..eq(5)


(x³ + y ³ )=(x+y) (x²+y²-xy)


183=9(x²+y²) -183/9

[From eq 3 & 5]

183 =( 9) (9(x²+y²)-183)/9

183 = (9(x²+y²)-183)

9(x²+y²)=183+183

9(x²+y²) =366


(x²+y²) =366/9……………….(6)


9/5 (a+b) (Given)

(9/5)[x² +y²]

(9/5)[366/9]

From equation 6

366/5 =73.2

Hence, the value of 9/5[a+b] = 73.2≈73


==================================================================

Hope this will help you...