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Question Number 11139 by Joel576 last updated on 13/Mar/17

Commented by Joel576 last updated on 13/Mar/17
$A real number t, so there is a three−ordered−pair solution {x, y, z} that fulfilled x^(2 ) + 2y^2 = 3z and x + y + z = t Find t Answer : t = −(9/8) 3t = 3x + 3y + 3z = 3x + 3y + x^2 + 2y^2 = (x + (3/2))^2 + 2(y + (3/4))^2 − ((27)/8) In order to have only a three−ordered−pair solution, so must be 3t = − ((27)/8) ⇔ t = − (9/8) Is there anyone who can explain that explanation above?$
$A real number t, so there is a three - ordered - pair$ $solution {x, y, z} that fulfilled$ $x^{2} + 2 y^{2} = 3 z and x + y + z = t$ $Find t$ $Answer : t = - \frac{9}{8}$ $3 t = 3 x + 3 y + 3 z$ $= 3 x + 3 y + x^{2} + 2 y^{2}$ $= {(x + \frac{3}{2})}^{2} + 2 {(y + \frac{3}{4})}^{2} - \frac{27}{8}$ $In order to have only a three - ordered - pair solution,$ $so must be 3 t = - \frac{27}{8} \Leftrightarrow t = - \frac{9}{8}$ $Is there anyone who can explain that explanation above ?$
	Answered by mrW1 last updated on 13/Mar/17

	$3 t = 3 x + 3 y + 3 z = 3 x + 3 y + x^{2} + 2 y^{2}$ $= x^{2} + 3 x + 2 (y^{2} + \frac{3}{2} y)$ $= {(x + \frac{3}{2})}^{2} + 2 {(y + \frac{3}{4})}^{2} - \frac{27}{8}$ ${(x + \frac{3}{2})}^{2} + 2 {(y + \frac{3}{4})}^{2} = 3 t + \frac{27}{8} = a$ $i f a > 0, o r t > - \frac{9}{8} :$ $t h e r e a r e i n f i n i t e s o l u t i o n s$ $f o r (x, y) . i n f a c t t h e e q u a t i o n$ ${(x + \frac{3}{2})}^{2} + 2 {(y + \frac{3}{4})}^{2} = a$ $d e s c r i b e s a n e l l i p s e . e a c h p o i n t o n i t$ $i s a s o l u t i o n .$ $i f a = 0, o r t = - \frac{9}{8} :$ $t h e r e i s o n l y o n e s o l u t i o n :$ ${(x + \frac{3}{2})}^{2} + 2 {(y + \frac{3}{4})}^{2} = 0$ $x = - \frac{3}{2} a n d y = - \frac{3}{4}$ $i f a < 0, o r t < - \frac{9}{8} :$ $t h e r e i s n o s o l u t i o n, s i n c e$ ${(x + \frac{3}{2})}^{2} + 2 {(y + \frac{3}{4})}^{2} ⩾ 0 ≮ 0$ $I f t h e r e i s o n e a n d o n l y o n e$ $s o l u t i o n f o r (x, y, z), t = - \frac{9}{8}$
	Commented by Joel576 last updated on 13/Mar/17

	$t h a n k y o u v e r y m u c h$
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