solve-inR-x-1-1-5-x-2-4x-9-1-10-2x-10-x-2-1-

Question Number 74880 by aliesam last updated on 03/Dec/19

s o l v e i n R

\sqrt[5]{∣ x + 1 ∣} - \sqrt[10]{x^{2} + 4 x - 9} = (2 x - 10) \sqrt{x^{2} + 1}

Answered by mind is power last updated on 03/Dec/19

x^{2} + 4 x - 9

x^{2} + 4 x - 9 > 0 \Rightarrow

x \in] - \infty, \frac{- 4 - \sqrt{52}}{2}] \cup] \frac{- 4 + \sqrt{52}}{2}, + \infty [

\sqrt[5]{∣ x + 1 ∣} ⩾ \sqrt[10]{x^{2} + 4 x - 9}

\Rightarrow x^{2} + 2 x + 1 ⩾ x^{2} + 4 x - 9

\Rightarrow 2 x ⩽ 10 \Rightarrow x ⩽ 5 \dots . h

let α solution

of

\sqrt[5]{∣ x + 1 ∣} - \sqrt[10]{x^{2} + 4 x - 9} = (2 x - 10) \sqrt{x^{2} + 1}

if α < 5

we get by . . H

\sqrt[5]{∣ α + 1 ∣} - \sqrt[10]{α^{2} + 4 α - 9} > 0

but (2 α - 10) \sqrt{α^{2} + 1} < 0 absurd

if α > 5

\sqrt[5]{∣ α + 1 ∣} - \sqrt[10]{α^{2} + 4 α - 9} < 0

(2 α - 10) \sqrt{α^{2} + 1} > 0 absurd

So if ther exist solution α, α \in R - {] - \infty, 5 [\cup] 5, + \infty [} \Leftrightarrow α \in {5}

for α = 5 we have

\sqrt[5]{6} - \sqrt[10]{36} = (2.5 - 10) \sqrt{26} = 0

true since 36 = 6^{2}

S = {5}