Solve ((1/x) − (1/x^3 ))^(1/2) + ((1/x^2 ) − (1/x^3 ))^(1/2) = 1


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Question Number 201991 by MATHEMATICSAM last updated on 18/Dec/23

$Solve$ ${(\frac{1}{x} - \frac{1}{x^{3}})}^{\frac{1}{2}} + {(\frac{1}{x^{2}} - \frac{1}{x^{3}})}^{\frac{1}{2}} = 1$
	Answered by mr W last updated on 18/Dec/23

	$\frac{\sqrt{\frac{1}{x}}}{1} = \frac{\sqrt{\frac{1}{x^{3}}}}{\sqrt{\frac{1}{x^{2}}}} = \sqrt{\frac{1}{x}}$ $\Rightarrow {(\sqrt{\frac{1}{x}})}^{2} + {(\sqrt{\frac{1}{x^{2}}})}^{2} = 1$ $\Rightarrow x^{2} - x - 1 = 0$ $\Rightarrow x = \frac{1 + \sqrt{5}}{2} ✓$
	Commented by mr W last updated on 19/Dec/23

	Answered by Rasheed.Sindhi last updated on 19/Dec/23
	$((1/x) − (1/x^3 ))^(1/2) + ((1/x^2 ) − (1/x^3 ))^(1/2) = 1 ((x/x^2 ) − (1/x^3 ))^(1/2) + ((1/x^2 ) − (1/x^3 ))^(1/2) = 1 (1/x){(x−(1/x))^(1/2) +(1−(1/x))^(1/2) }=1 (x−(1/x))^(1/2) +(1−(1/x))^(1/2) =x Let (√(x−(1/x))) =a , (√(1−(1/x))) =b a+b=x ∧ a^2 −b^2 =x−1 ⇒a−b=((a^2 −b^2 )/(a+b))=((x−1)/x) { (( a+b=x)),((a−b=((x−1)/x)=1−(1/x))) :} ⇒ { ((2a=x+((x−1)/x)=x−(1/x)+1=a^2 +1)),((2b=x−((x−1)/x)=x+(1/x)−1=x−(1−(1/x))=x−b^2 )) :} ⇒ { ((a^2 −2a+1=0⇒(a−1)^2 =0⇒a=1⇒b^2 +b=a=1)),((b^2 +2b=x)) :} ⇒(√(x−(1/x))) =1⇒x−(1/x)=1⇒x^2 −x−1=0 ⇒x_(≥0) =((1+(√(1+4)))/2)=((1+(√5) )/2)$
	${(\frac{1}{x} - \frac{1}{x^{3}})}^{\frac{1}{2}} + {(\frac{1}{x^{2}} - \frac{1}{x^{3}})}^{\frac{1}{2}} = 1$ ${(\frac{x}{x^{2}} - \frac{1}{x^{3}})}^{\frac{1}{2}} + {(\frac{1}{x^{2}} - \frac{1}{x^{3}})}^{\frac{1}{2}} = 1$ $\frac{1}{x} {{(x - \frac{1}{x})}^{1 / 2} + {(1 - \frac{1}{x})}^{1 / 2}} = 1$ ${(x - \frac{1}{x})}^{1 / 2} + {(1 - \frac{1}{x})}^{1 / 2} = x$ $L e t \sqrt{x - \frac{1}{x}} = a, \sqrt{1 - \frac{1}{x}} = b$ $a + b = x \land a^{2} - b^{2} = x - 1$ $\Rightarrow a - b = \frac{a^{2} - b^{2}}{a + b} = \frac{x - 1}{x}$ ${\begin{cases} a + b = x \\ a - b = \frac{x - 1}{x} = 1 - \frac{1}{x} \end{cases}$ $\Rightarrow {\begin{cases} 2 a = x + \frac{x - 1}{x} = x - \frac{1}{x} + 1 = a^{2} + 1 \\ 2 b = x - \frac{x - 1}{x} = x + \frac{1}{x} - 1 = x - (1 - \frac{1}{x}) = x - b^{2} \end{cases}$ $\Rightarrow {\begin{cases} a^{2} - 2 a + 1 = 0 \Rightarrow {(a - 1)}^{2} = 0 \Rightarrow a = 1 \Rightarrow b^{2} + b = a = 1 \\ b^{2} + 2 b = x \end{cases}$ $\Rightarrow \sqrt{x - \frac{1}{x}} = 1 \Rightarrow x - \frac{1}{x} = 1 \Rightarrow x^{2} - x - 1 = 0$ $\Rightarrow x_{⩾ 0} = \frac{1 + \sqrt{1 + 4}}{2} = \frac{1 + \sqrt{5}}{2}$
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