﻿ If a,b\in R such that a\sqrt{a} +b\sqrt{b} =183 and b\sqrt{a} +a\sqrt{b} =182, find the value

#### Question

If a,b\in R such that a\sqrt{a} +b\sqrt{b} =183 and b\sqrt{a} +a\sqrt{b} =182, find the value of \dfrac{9}{5}(a+b)

Collected in the board: Algebraic equation

Steven Zheng posted 4 months ago

Given conditions,

a\sqrt{a} +b\sqrt{b} =183
(1)
b\sqrt{a} +a\sqrt{b} =182
(2)

Let

x=\sqrt{a}, y=\sqrt{b}

That is

a=x^2, b=y^2

Given conditions (1) and (2) are converted to

x^3+y^3 = 183
(3)
xy^2+x^2y = 182
(4)

Apply the formula for the cube of a binomial,

(x+y)^3 = x^3+3x^2y+3xy^2+y^3

=(x^3+y^3)+3(xy^2+x^2y)

=183+3\times 182

=729=9^3

Obtain the value of x+y by taking the cube root of both sides of the equation:

x+y = 9
(5)

Obtain the value of xy by factoring out xy from equation (4)

xy(x+y) = 182

xy = \dfrac{182}{x+y}

=\dfrac{182}{9}
(6)

Square both sides of equation (5)

(x+y)^2 = 81

x^2+y^2+2xy = 81

Substitute the value of xy

x^2+y^2 = 81-2\cdotp \dfrac{182}{9} =\dfrac{365}{9}

Therefore,

a+b = \dfrac{365}{9}

Now we can get the final value

\dfrac{9}{5}(a+b) =\dfrac{9}{5}\cdotp \dfrac{365}{9} =73

Steven Zheng posted 4 months ago

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