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Question Number 62263 by Tawa1 last updated on 18/Jun/19

	Answered by behi83417@gmail.com last updated on 19/Jun/19
	$x+(√(x−2))=a⇒(2/a)=(1/(x+(√(x−2))))=((x−(√(x−2)))/1) (2/a)+(√a)=3⇒a(√a)−3a+2=0 (√a)=t⇒t^3 −3t^2 +2=0 ⇒(t−1)(t^2 +mt+n)=t^3 −3t^2 +2 t=0⇒−n=2⇒n=−2 t=−1⇒−2(1−m−2)=−1−3+2 ⇒−m−1=1⇒m=−2 ⇒(t−1)(t^2 −2t−2)=0 1. { ((t=1⇒(√a)=1⇒a=1⇒x+(√(x−2))=1)),((⇒x−2=1−2x+x^2 ⇒x^2 −3x+3=0)) :} ⇒x=((3±(√(9−12)))/2)=(1/2)(3±i(√3)) 2. { ((t^2 −2t−2=0⇒(t−1)^2 =3⇒t=1±(√3))),((a=(1±(√3))^2 )) :} x+(√(x−2))=a⇒x−2=a^2 −2ax+x^2 ⇒x^2 −(2a+1)x+a^2 +2=0 x=((2a+1±(√(4a^2 +4a+1−4a^2 −8)))/2) x=((2a+1±(√(4a−7)))/2)=(1/2)[9±2(√3)±(√(9±8(√3)))] 4a−7=4(4±2(√3))−7=9±8(√3) 2a+1=2(4±(√3))+1=9±2(√3)$
	$x + \sqrt{x - 2} = a \Rightarrow \frac{2}{a} = \frac{1}{x + \sqrt{x - 2}} = \frac{x - \sqrt{x - 2}}{1}$ $\frac{2}{a} + \sqrt{a} = 3 \Rightarrow a \sqrt{a} - 3 a + 2 = 0$ $\sqrt{a} = t \Rightarrow t^{3} - {3 t}^{2} + 2 = 0$ $\Rightarrow (t - 1) (t^{2} + mt + n) = t^{3} - {3 t}^{2} + 2$ $t = 0 \Rightarrow - n = 2 \Rightarrow n = - 2$ $t = - 1 \Rightarrow - 2 (1 - m - 2) = - 1 - 3 + 2$ $\Rightarrow - m - 1 = 1 \Rightarrow m = - 2$ $\Rightarrow (t - 1) (t^{2} - 2 t - 2) = 0$ $1 . {\begin{cases} t = 1 \Rightarrow \sqrt{a} = 1 \Rightarrow a = 1 \Rightarrow x + \sqrt{x - 2} = 1 \\ \Rightarrow x - 2 = 1 - 2 x + x^{2} \Rightarrow x^{2} - 3 x + 3 = 0 \end{cases}$ $\Rightarrow x = \frac{3 \pm \sqrt{9 - 12}}{2} = \frac{1}{2} (3 \pm i \sqrt{3})$ $2 . {\begin{cases} t^{2} - 2 t - 2 = 0 \Rightarrow {(t - 1)}^{2} = 3 \Rightarrow t = 1 \pm \sqrt{3} \\ a = {(1 \pm \sqrt{3})}^{2} \end{cases}$ $x + \sqrt{x - 2} = a \Rightarrow x - 2 = a^{2} - 2 ax + x^{2}$ $\Rightarrow x^{2} - (2 a + 1) x + a^{2} + 2 = 0$ $x = \frac{2 a + 1 \pm \sqrt{{4 a}^{2} + 4 a + 1 - {4 a}^{2} - 8}}{2}$ $x = \frac{2 a + 1 \pm \sqrt{4 a - 7}}{2} = \frac{1}{2} [9 \pm 2 \sqrt{3} \pm \sqrt{9 \pm 8 \sqrt{3}}]$ $4 a - 7 = 4 (4 \pm 2 \sqrt{3}) - 7 = 9 \pm 8 \sqrt{3}$ $2 a + 1 = 2 (4 \pm \sqrt{3}) + 1 = 9 \pm 2 \sqrt{3}$
	Commented by Tawa1 last updated on 19/Jun/19

	$God bless you sir$
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