x^3 +x−1=^3 (√(2x^3 +11))+(√(5x^2 +16)) Find x∈R


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 143684 by Huy last updated on 17/Jun/21

$x^{3} + x - 1 =^{3} \sqrt{{2 x}^{3} + 11} + \sqrt{{5 x}^{2} + 16}$ $Find x \in R$
	Answered by TheHoneyCat last updated on 17/Jun/21

	$let x_{0} = 2$ $x_{0} is a solution (2^{3} + 2 - 1 = 9 =^{3} \sqrt{27} + \sqrt{36})$ $Juging by the graph, it seems it is the only solution$ $let me proove it$ $let x \in] x_{0}, + \infty [$ $let f : x \to x^{3} + x - 1$ $and g : x \to^{3} \sqrt{{2 x}^{3} + 11} + \sqrt{{5 x}^{2} + 16}$ $\frac{d f}{d x} = 3 x^{2} + 1$ $\frac{d g}{d x} = 6 x^{2} {(2 x^{3} + 11)}^{- \frac{2}{3}} + 10 x {(5 x^{2} + 16)}^{- \frac{1}{2}}$ $x ⩾ 2$ $\Rightarrow 2 x^{3} ⩾ 16$ $\Rightarrow 2 x^{3} + 11 ⩾ 27$ $\Rightarrow {(2 x^{3} + 11)}^{\frac{1}{3}} ⩾ 3$ $\Rightarrow {(2 x^{3} + 11)}^{\frac{2}{3}} ⩾ 9$ $\Rightarrow {(2 x^{3} + 11)}^{\frac{- 2}{3}} ⩽ \frac{1}{9}$ $\Rightarrow 6 x^{2} {(2 x^{3} + 11)}^{\frac{- 2}{3}} ⩽ \frac{2}{9} 3 x^{2}$ $5 x^{2} + 16 ⩾ 5 x^{2}$ $\Rightarrow {(5 x^{2} + 16)}^{\frac{1}{2}} ⩾ \sqrt{5} x$ $\Rightarrow 10 x {(5 x^{2} + 16)}^{\frac{- 1}{2}} ⩽ \frac{10}{\sqrt{5}}$ $and 3 x^{2} + 1 ⩾ \frac{2}{9} 3 x^{2} + \frac{10}{\sqrt{5}}$ $so \frac{d f}{d x} ⩾ \frac{d g}{d x}$ $so \forall x ⩾ x_{0} f (x) \neq g (x)$ $t h e s a m e r e a s o n n i n g c a n b e d o n n e f o r$ $x \in [0, x_{0}]$ $and for x < 0 checking the sine is {sufficient}_{◼}$
	Commented by TheHoneyCat last updated on 17/Jun/21


Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum