x-2-x-a-x-a-a-R-solve-for-x-

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

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

\dots \dots \dots

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 . .

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