(x^2 /(x+a))+(√x)=a (a∈R) solve for: x .


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 90194 by behi83417@gmail.com last updated on 21/Apr/20

$\frac{x^{2}}{x + a} + \sqrt{x} = a (a \in R)$ $solve for : x .$
	Answered by ajfour last updated on 22/Apr/20
	$let (√x)=t t^4 =(a−t)(t^2 +a) t^4 +t^3 −at^2 +at−a^2 =0 let (t^2 +pt+q)(t^2 +rt+m)=0 ⇒ t^4 +(p+r)t^3 +(m+pr+q)t^2 +(pm+qr)t+qm=0 ⇒ p+r=1 , pr+m+q=−a, pm+qr =a , qm=−a^2 ⇒ pm−((a^2 r)/m)=a pm^2 −am−a^2 r=0 m=(a/(2p))±(√((a^2 /(4p^2 ))+((a^2 r)/p))) m= (a/(2p)){1±(√(1+4pr))} q=−(a^2 /m) = (2ap)(({1∓(√(1+4pr))})/(4pr)) q= (a/(2r)){1∓(√(1+4pr))} q+m=(a/(2p))+(a/(2r))±(√(1+4pr))((a/(2p))−(a/(2r))) −a−pr=(a/2)(((p+r)/(pr)))±(a/2)(((r−p)/(pr)))(√(1+4pr)) let pr=z −a−z=(a/(2z))±(a/(2z))(√(1−16z^2 )) ⇒ ((a/(2z))+z+a)^2 =(a^2 /(4z^2 ))(1−16z^2 ) z^2 +a^2 +a+2az+(a^2 /z)+4a^2 =0 z^3 +2az^2 +a(5a+1)z+a^2 =0 let z=y−((2a)/3) y^3 −2ay^2 +((4a^2 y)/3)−((8a^3 )/(27)) +2ay^2 −((8a^2 y)/3)+((8a^3 )/9) +5a^2 y+ay−((10a^3 )/3)−((2a^2 )/3)+a^2 =0 y^3 +(((11a^2 )/3)+a)y−((70a^3 )/(27))+(a^2 /3)=0 .........$
	$l e t \sqrt{x} = t$ $t^{4} = (a - t) (t^{2} + a)$ $t^{4} + t^{3} - {a t}^{2} + a t - a^{2} = 0$ $l e t$ $(t^{2} + p t + q) (t^{2} + r t + m) = 0$ $\Rightarrow$ $t^{4} + (p + r) t^{3} + (m + p r + q) t^{2}$ $+ (p m + q r) t + q m = 0$ $\Rightarrow p + r = 1, p r + m + q = - a,$ $p m + q r = a, q m = - a^{2}$ $\Rightarrow p m - \frac{a^{2} r}{m} = a$ ${p m}^{2} - a m - a^{2} r = 0$ $m = \frac{a}{2 p} \pm \sqrt{\frac{a^{2}}{4 p^{2}} + \frac{a^{2} r}{p}}$ $m = \frac{a}{2 p} {1 \pm \sqrt{1 + 4 p r}}$ $q = - \frac{a^{2}}{m} = (2 a p) \frac{{1 \mp \sqrt{1 + 4 p r}}}{4 p r}$ $q = \frac{a}{2 r} {1 \mp \sqrt{1 + 4 p r}}$ $q + m = \frac{a}{2 p} + \frac{a}{2 r} \pm \sqrt{1 + 4 p r} (\frac{a}{2 p} - \frac{a}{2 r})$ $- a - p r = \frac{a}{2} (\frac{p + r}{p r}) \pm \frac{a}{2} (\frac{r - p}{p r}) \sqrt{1 + 4 p r}$ $l e t p r = z$ $- a - z = \frac{a}{2 z} \pm \frac{a}{2 z} \sqrt{1 - 16 z^{2}}$ $\Rightarrow {(\frac{a}{2 z} + z + a)}^{2} = \frac{a^{2}}{4 z^{2}} (1 - 16 z^{2})$ $z^{2} + a^{2} + a + 2 a z + \frac{a^{2}}{z} + 4 a^{2} = 0$ $z^{3} + 2 {a z}^{2} + a (5 a + 1) z + a^{2} = 0$ $l e t z = y - \frac{2 a}{3}$ $y^{3} - 2 {a y}^{2} + \frac{4 a^{2} y}{3} - \frac{8 a^{3}}{27}$ $+ 2 {a y}^{2} - \frac{8 a^{2} y}{3} + \frac{8 a^{3}}{9}$ $+ 5 a^{2} y + a y - \frac{10 a^{3}}{3} - \frac{2 a^{2}}{3} + a^{2} = 0$ $y^{3} + (\frac{11 a^{2}}{3} + a) y - \frac{70 a^{3}}{27} + \frac{a^{2}}{3} = 0$ $. . . . . . . . .$
	Commented by behi83417@gmail.com last updated on 22/Apr/20

	$thank you very much dear Ajfour .$
	Commented by ajfour last updated on 22/Apr/20

	$i w i l l t r y t o f i n i s h . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum