Change in Q#27507 Solve simultaneously: 2(√x)+y=13 x+2(√y)=10


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Question Number 27559 by Rasheed.Sindhi last updated on 09/Jan/18

$You can't use 'macro parameter character #' in math mode$ $Solve simultaneously :$ $2 \sqrt{x} + y = 13$ $x + 2 \sqrt{y} = 10$
Commented by Rasheed.Sindhi last updated on 09/Jan/18

$x = 4, y = 9 is solution$ $Method of solving is required .$
	Answered by Amstrongmazoka last updated on 09/Jan/18

	$F r o m e q u a t i o n (1) y = 13 - 2 \sqrt{x}$ $\Rightarrow s u b s t i t u t i n g f o r y i n e q u a t i o n (2), g i v e s,$ $x + 2 \sqrt{(13 - 2 \sqrt{x})} = 10 ∴ \sqrt{(13 - 2 \sqrt{x})} = 5 - \frac{x}{2}$ $N o w s q u a r i n g a b o v e s i d e s, g i v e s$ $13 - 2 \sqrt{x} = {(5 - \frac{x}{2})}^{2} \Rightarrow 13 - 2 \sqrt{x} = 25 - 5 x + \frac{x^{2}}{4}$ $∴ - 2 \sqrt{x} = 12 - 5 x + \frac{x^{2}}{4} . A g a i n, s q u a r i n g b o t h s i d e s g i v e s$ $4 x = 144 - 120 x + 25 x^{2} + 6 x^{2} - \frac{5 x^{3}}{2} + \frac{x^{4}}{16} .$ $G r o u p i n g l i k e t e r m s g i v e s$ $\frac{x^{4}}{16} - \frac{5 x^{3}}{2} + 31 x^{2} - 124 x + 144 = 0 . M u l t i p l y i n g t h r o u g h b y 16 g i v e s$ $x^{4} - 40 x^{3} + 496 x^{2} - 1984 x + 2304 = 0 . T h i s i s a p o l y n o m i a l o f d e g r e e 4 a n d i t c a n b e s o l v e d u s i n g f a c t o r i s a t i o n o r {N e w t o n}^{'} m e t h o d .$ $L e t f (x) = x^{4} - 40 x^{3} + 496 x^{2} - 1984 x + 2304$ $\Rightarrow f (4) = 4^{4} - 40 (4^{3}) + 496 (4^{2}) - 1984 (4) + 2304 = 0$ $B y t h e f a c t o r t h e o r e m s i n c e f (4) = 0, \Rightarrow (x - 4) i s a f a c t o r o f f (x) .$ $T h e o t h e r f a c t o r s a r e f o u n d a s f o l l o w s :$ $x^{3} - 36 x^{2} + 352 x - 576$ $(x - 4) \sqrt{x^{4} - 40 x^{3} + 496 x^{2} - 1984 x + 2304}$ $- (x^{4} - 4 x^{3})$ $- 36 x^{3} + 496 x^{2}$ $- (- 36 x^{3} + 144 x^{2})$ $352 x^{2} - 1984 x$ $- (352 x^{2} - 1408 x)$ $- 576 x + 2304$ $- (- 576 x + 2304)$ $- - - - - - -$ $∴ f (x) = (x - 4) (x^{3} - 36 x^{2} + 352 x - 576) \Rightarrow x = 4, x = 2.0365, x = 19.337$ $B u t y = 13 - 2 \sqrt{x}, \Rightarrow w h e n x = 4, y = 13 - 2 \sqrt{4} = 9, w h e n x = 2.0365, y = 13 - 2 \sqrt{2.0365} = 10.1459 a n d w h e n x = 19.337, y = 13 - 2 \sqrt{19.337} = 4.2052$ $t h u s (x = 4, y = 9), (x = 2.0365, y = 10.1459) a n d (x = 19.337, y = 4.2052)$ $a r e t h e p o s s i b l e p a i r s f r o m w h i c h a s o l u t i o n c a n b e f o u n d .$ $a m o u n t t h e m, o n l y t h e f i r s t p a i r s a t i s f y t h e e q u a t i o n s s i m u l t a n e o u s l y .$ $∴ x = 4, y = 9$
	Commented by Rasheed.Sindhi last updated on 09/Jan/18

	$In finding factors trial method$ $is involved .$ $A n y W a y T h a n X - a - l o t!$
	Commented by Rasheed.Sindhi last updated on 09/Jan/18

	$What if we are only sure of$ $real roots ?$
	Answered by Rasheed.Sindhi last updated on 10/Jan/18

	$Trying to find solution in$ $whole numbers .$ $2 \sqrt{x} + y = 13 . . . . . . . . (i)$ $x + 2 \sqrt{y} = 10 . . . . . . . . (ii)$ $(i) \Rightarrow \sqrt{x} = \frac{13 - y}{2} . . . . . . . . . . (iii)$ $(ii) \Rightarrow x = 10 - 2 \sqrt{y}$ $\sqrt{x} = \sqrt{10 - 2 \sqrt{y}} . . . . . . (iv)$ $(iii) & (iv) \Rightarrow \frac{13 - y}{2} = \sqrt{10 - 2 \sqrt{y}}$ $\sqrt{p - q \sqrt{c}} can be changed into$ $a + b \sqrt{c} form in some cases and$ $the process is as under .$ $Let \sqrt{10 - 2 \sqrt{y}} = a + b \sqrt{y} a, b \in Q$ ${(\sqrt{10 - 2 \sqrt{y}})}^{2} = {(a + b \sqrt{y})}^{2}$ $a^{2} + b^{2} y + 2 ab \sqrt{y} = 10 - 2 \sqrt{y}$ $By comparing the coefficients$ $of \sqrt{y} and the terms not containing$ $\sqrt{y} {we}^{'} ll get :$ $a^{2} + b^{2} y = 10 \land 2 ab = - 2$ $ab = - 1$ $b = - 1 / a$ $a^{2} + {(- 1 / a)}^{2} y = 10$ $a^{4} - {10 a}^{2} + y = 0$ $a^{2} = \frac{10 \pm \sqrt{100 - 4 y}}{2}$ $\frac{10 \pm \sqrt{100 - 4 y}}{2} is perfect square$ $rational number \Rightarrow 100 - 4 y is$ $perfect square rational number$ $and non - negative .$ $∴ y = 0, 9, 16, 21$ $y = 0 \Rightarrow a = \frac{10 \pm 10}{2} = \overset{\times}{\sqrt{10}}, 0$ $a \neq 0 because in that case b = - 1 / 0 = \infty$ $y = 9 \Rightarrow a = \frac{10 \pm 8}{2} = 3, 1$ $y = 16 \Rightarrow a = \frac{10 \pm 6}{2} = \overset{\times}{\sqrt{8}}, \overset{\times}{\sqrt{2}}$ $y = 21 \Rightarrow a = \frac{10 \pm 4}{2} = \overset{\times}{\sqrt{7}}, \overset{\times}{\sqrt{3}}$ $So y may be 9 and a = 1, 3$ $a = 3, b = - 1 / 3 & y = 9 satisfy the$ $following$ $\sqrt{10 - 2 \sqrt{y}} = a + b \sqrt{y}$ $\sqrt{10 - 2 \sqrt{9}} = 3 - \frac{1}{3} \sqrt{9}$ $2 = 2$ $Hence y = 9$ $Continue$
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