Solve-in-R-a-3-x-3-8-x-3-4-0-b-x-2-x-6-x-1-c-x-3-27-6-lt-x-3-

Question Number 104178 by mathocean1 last updated on 19/Jul/20

S o l v e i n R

a) 3 ∣ x - \sqrt{3} ∣ - 8 \sqrt{(∣ x - \sqrt{3} ∣) + 4} = 0

b) \sqrt{(∣ x^{2} - x - 6} ∣) = x + 1

c) \sqrt{(x^{3} - 27} + 6 < x + 3

Answered by OlafThorendsen last updated on 19/Jul/20

a) u = ∣ x - \sqrt{3} ∣

9 u^{2} = 64 (u + 4)

9 u^{2} - 64 u - 256 = 0

Δ = 13312

u = \frac{64 \pm 32 \sqrt{13}}{18} = \frac{16}{9} (2 \pm \sqrt{13})

u > 0 \Rightarrow u = \frac{16}{9} (2 + \sqrt{13})

x - \sqrt{3} = \pm \frac{16}{9} (2 + \sqrt{13})

x = \sqrt{3} \pm \frac{16}{9} (2 + \sqrt{13})

Commented by mathocean1 last updated on 19/Jul/20

T h a n k y o u s i r!

Answered by bemath last updated on 20/Jul/20

(c) \sqrt{x^{3} - 27} < x - 3

(1) x - 3 > 0 \to x > 3

(2) x^{3} - 27 ⩾ 0 \Rightarrow x ⩾ 3

(3) x^{3} - 27 < x^{2} - 6 x + 9

(x - 3) (x^{2} + 3 x + 9) < {(x - 3)}^{2}

(x - 3) {x^{2} + 3 x + 9 + 3 - x} < 0

(x - 3) (x^{2} + 2 x + 12) < 0

(x - 3) ({(x + 1)}^{2} + 11) < 0

x < 3

s o l u t i o n (1) \cap (2) \cap (3)

x = \emptyset

Commented by mathocean1 last updated on 20/Jul/20

t h a n k y o u s i r

Answered by Rasheed.Sindhi last updated on 20/Jul/20

a) 3 ∣ x - \sqrt{3} ∣ - 8 \sqrt{(∣ x - \sqrt{3} ∣) + 4} = 0

L e t \sqrt{(∣ x - \sqrt{3} ∣) + 4} = y

∣ x - \sqrt{3} ∣= y^{2} - 4

3 (y^{2} - 4) - 8 y = 0

3 y^{2} - 8 y - 12 = 0

y = \frac{8 \pm \sqrt{64 + 144}}{3}

\sqrt{(∣ x - \sqrt{3} ∣) + 4} = \frac{8 \pm 4 \sqrt{13}}{3}

(∣ x - \sqrt{3} ∣) + 4 = {(\frac{8 \pm 4 \sqrt{13}}{3})}^{2}

∣ x - \sqrt{3} ∣= {(\frac{8 \pm 4 \sqrt{13}}{3})}^{2} - 4

∣ x - \sqrt{3} ∣= {(\frac{8 \pm 4 \sqrt{13}}{3})}^{2} - 4

{\begin{cases} x - \sqrt{3} = {(\frac{8 \pm 4 \sqrt{13}}{3})}^{2} - 4 > 0 \\ - x + \sqrt{3} = {(\frac{8 \pm 4 \sqrt{13}}{3})}^{2} - 4 > 0 \end{cases}

{\begin{cases} x = {(\frac{8 \pm 4 \sqrt{13}}{3})}^{2} - 4 + \sqrt{3} \\ x = - {(\frac{8 \pm 4 \sqrt{13}}{3})}^{2} + 4 + \sqrt{3} \end{cases}