a-x-b-y-2-ab-x-a-y-b-2-ab-a-b-R-

Question Number 62275 by behi83417@gmail.com last updated on 18/Jun/19

{\begin{cases} a \sqrt{x} + b \sqrt{y} = 2 \sqrt{ab} \\ x \sqrt{a} + y \sqrt{b} = 2 \sqrt{ab} \end{cases} a, b \in R^{+}

Answered by mr W last updated on 19/Jun/19

l e t X = \sqrt{x}, Y = \sqrt{y}, A = \sqrt{a}, B = \sqrt{b}

A^{2} X + B^{2} Y = 2 A B

\Rightarrow Y = \frac{2 A B - A^{2} X}{B^{2}}

{A X}^{2} + {B Y}^{2} = 2 A B

{A X}^{2} + B \frac{{(2 A B - A^{2} X)}^{2}}{B^{4}} = 2 A B

(A^{3} + B^{3}) X^{2} - 4 A^{2} B X + 2 B^{2} (2 A - B^{2}) = 0

\Rightarrow X = \frac{2 A^{2} B \pm B^{2} \sqrt{2 (A^{3} + B^{3}) - 4 A B}}{A^{3} + B^{3}}

\Rightarrow x = {\frac{2 a \sqrt{b} \pm b \sqrt{2 (a \sqrt{a} + b \sqrt{b}) - 4 \sqrt{a b}}}{a \sqrt{a} + b \sqrt{b}}}^{2}

s i m i l a r l y

(A^{3} + B^{3}) Y^{2} - 4 {A B}^{2} Y + 2 A^{2} (2 B - A^{2}) = 0

\Rightarrow Y = \frac{2 {A B}^{2} \pm A^{2} \sqrt{2 (A^{3} + B^{2}) - 4 A B}}{A^{3} + B^{3}}

\Rightarrow y = {\frac{2 \sqrt{a} b \pm a \sqrt{2 (a \sqrt{a} + b \sqrt{b}) - 4 \sqrt{a b}}}{a \sqrt{a} + b \sqrt{b}}}^{2}

Commented by behi83417@gmail.com last updated on 20/Jun/19

perfect sir . thank you so much dear master .