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Question Number 81674 by Power last updated on 14/Feb/20

Commented by Tony Lin last updated on 14/Feb/20
$let y=7−x^2 x+(7−x^2 )^2 =11 x+(x^4 −14x^2 +49)−11=0 x^4 −14x^2 +x+38=0 (x−2)(x^3 +2x^2 −10x−19)=0 x^3 +2x^2 −10x−19=0 ⇒ { ((x_1 ≈−3.2832, y_1 ≈−3.7794)),((x_2 ≈−1.8481, y_2 ≈3.5845)),((x_3 ≈3.1313, y_3 ≈−2.8050)) :} (you can use IVT to evaluate the value) of course x_4 =2 and y_4 =3$
$l e t y = 7 - x^{2}$ $x + {(7 - x^{2})}^{2} = 11$ $x + (x^{4} - 14 x^{2} + 49) - 11 = 0$ $x^{4} - 14 x^{2} + x + 38 = 0$ $(x - 2) (x^{3} + 2 x^{2} - 10 x - 19) = 0$ $x^{3} + 2 x^{2} - 10 x - 19 = 0$ $\Rightarrow {\begin{cases} x_{1} \approx - 3.2832, y_{1} \approx - 3.7794 \\ x_{2} \approx - 1.8481, y_{2} \approx 3.5845 \\ x_{3} \approx 3.1313, y_{3} \approx - 2.8050 \end{cases}$ $(y o u c a n u s e I V T t o e v a l u a t e t h e v a l u e)$ $o f c o u r s e x_{4} = 2 a n d y_{4} = 3$
Commented by mr W last updated on 14/Feb/20

$y = 7 - x^{2}$ $x + {(7 - x^{2})}^{2} = 11$ $x^{4} - 14 x^{2} + x + 38 = 0$ $(x - 2) (x^{3} + 2 x^{2} - 10 x - 19) = 0$ $x - 2 = 0 \Rightarrow x = 2 \Rightarrow y = 3$ $x^{3} + 2 x^{2} - 10 x - 19 = 0$ $l e t x = s - \frac{2}{3}$ $\Rightarrow s^{3} - \frac{34}{3} s - \frac{317}{27} = 0$ $Δ = {(- \frac{34}{9})}^{3} + {(- \frac{317}{54})}^{2} < 0$ $s = \frac{2 \sqrt{34}}{3} \sin (- \frac{1}{3} \sin^{- 1} \frac{317 \sqrt{34}}{2312} + \frac{2 k π}{3})$ $\Rightarrow x = - \frac{2}{3} + \frac{2 \sqrt{34}}{3} \sin (- \frac{1}{3} \sin^{- 1} \frac{317 \sqrt{34}}{2312} + \frac{2 k π}{3})$ $\Rightarrow x = - 1.848126 \Rightarrow y = 3.584427$ $\Rightarrow x = 3.131313 \Rightarrow y = - 2.805118$ $\Rightarrow x = - 3.283186 \Rightarrow y = - 3.779319$
Commented by Power last updated on 14/Feb/20

$thanks$
	Answered by Prithwish Sen 1 last updated on 14/Feb/20

	$x = 2$ $y = 3$
	Commented by Power last updated on 14/Feb/20

	$solve sir proved$
	Answered by behi83417@gmail.com last updated on 14/Feb/20
	$(x^2 −4)+(y−3)=0 (x−2)+(y^2 −9)=0 ⇒ { (((x−2)(x+2)+(y−3)=0)),(((x−2)+(y−3)(y+3)=0)) :} ⇒ { (((x−2)−(x−2)(x+2)(y+3)=0)),((⇒(x−2)[1−(x+2)(y+3)]=0)) :} ⇒[x−2=0,y^2 =11−2=9⇒y=±3]$
	$(x^{2} - 4) + (y - 3) = 0$ $(x - 2) + (y^{2} - 9) = 0$ $\Rightarrow {\begin{cases} (x - 2) (x + 2) + (y - 3) = 0 \\ (x - 2) + (y - 3) (y + 3) = 0 \end{cases}$ $\Rightarrow {\begin{cases} (x - 2) - (x - 2) (x + 2) (y + 3) = 0 \\ \Rightarrow (x - 2) [1 - (x + 2) (y + 3)] = 0 \end{cases}$ $\Rightarrow [x - 2 = 0, y^{2} = 11 - 2 = 9 \Rightarrow y = \pm 3]$
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