2-x-y-9-1-x-2-y-3-2-solve-the-simultaneous-equation-

Question Number 27507 by NECx last updated on 07/Jan/18

2 \sqrt{x} + y = 9 \dots . (1)

x + 2 \sqrt{y} = 3 \dots . (2)

s o l v e t h e s i m u l t a n e o u s e q u a t i o n

Commented by Rasheed.Sindhi last updated on 08/Jan/18

(1) + (2) :

x + y + 2 (\sqrt{x} + \sqrt{y}) = 12 \dots . . (3)

(1) - (2) :

- (x - y) + 2 (\sqrt{x} - \sqrt{y}) = 6 \dots (4)

Let \sqrt{x} + \sqrt{y} = u & \sqrt{x} - \sqrt{y} = v

{(a + b)}^{2} - {(a - b)}^{2} = 4 ab

u^{2} - v^{2} = 4 \sqrt{xy}

^{∙} \sqrt{xy} = \frac{u^{2} - v^{2}}{4}

{(\sqrt{x} + \sqrt{y})}^{2} = u^{2}

x + y + 2 \sqrt{xy} = u^{2}

^{∙} x + y = u^{2} - 2 (\frac{u^{2} - v^{2}}{4})

= \frac{u^{2} + v^{2}}{2}

{(a - b)}^{2} = {(a + b)}^{2} - 4 ab

^{∙} {(x - y)}^{2} = {(\frac{u^{2} + v^{2}}{2})}^{2} - 4 {(\frac{u^{2} - v^{2}}{4})}^{2}

= \frac{{(u^{2} + v^{2})}^{2}}{4} - \frac{{(u^{2} - v^{2})}^{2}}{4}

= \frac{u^{4} + {2 u}^{2} v^{2} + v^{4} - u^{4} + {2 u}^{2} v^{2} - v^{4}}{4}

{(x - y)}^{2} = u^{2} v^{2}

x - y = \pm uv

(3) : x + y + 2 (\sqrt{x} + \sqrt{y}) = 12

\frac{u^{2} + v^{2}}{2} + 2 u = 12

u^{2} + 4 u + v^{2} = 24 \dots \dots . (5)

(4) : - (x - y) + 2 (\sqrt{x} - \sqrt{y}) = 6

- (\pm uv) + 2 v = 6

\mp uv + 2 v = 6

v = \frac{6}{\mp u + 2}

(5) : u^{2} + 4 u + {(\frac{6}{\mp u + 2})}^{2} = 24

(5) : u (u + 4) + \frac{36}{{(\mp u + 2)}^{2}} = 24

u (u + 4) {(\mp u + 2)}^{2} + 36 = 24 {(\mp u + 2)}^{2}

Continue

Commented by prakash jain last updated on 08/Jan/18

2 \sqrt{x} + y = 9

y_{1} = 9 - 2 \sqrt{x}

x + 2 \sqrt{y} = 3

2 \sqrt{y} = 3 - x

for real y, x ⩽ 3

y_{2} = \frac{1}{4} {(3 - x)}^{2}

u = y_{1} - y_{2} = 9 - 2 \sqrt{x} - \frac{1}{4} {(3 - x)}^{2}

f o r r e a l s o l u t i o n x ⩽ 3

x ⩽ 3 \Rightarrow 9 - 2 \sqrt{x} ⩾ 9 - 2 \sqrt{3} > 5

\frac{1}{4} {(3 - x)}^{2} ⩽ \frac{9}{4}

u ⩾ 5 - \frac{9}{4} > 0 \Rightarrow n o r e a l s o l u t i o n

Commented by Rasheed.Sindhi last updated on 08/Jan/18

New way for me to attack such

problem!

Commented by prakash jain last updated on 08/Jan/18

Actually finding the complex solutions is tough.

Answered by jota@ last updated on 09/Jan/18

t h e r e i s n o t a r e a l s o l u t i o n .

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