Question

$\text{If } x=\sqrt{\dfrac{\sqrt{10}+1 }{\sqrt{10}-1 } }, \ \text{find the value of } x-\dfrac{1}{x}$

Answer 1

Topic

• Irrational Numbers- Rationalization

It is a process of converting the denominator to a rational number.

• Polynomials- Quadratic Equation

A quadratic equation has the highest degree of 2.

Solution

The following is the given value of $x$.

$x=\sqrt{\dfrac{\sqrt{10}+1}{\sqrt{10} -1} }$

$\implies x=\sqrt{\dfrac{\sqrt{10}+1}{\sqrt{10} -1} }\times \sqrt{\dfrac{\sqrt{10}+1}{\sqrt{10} +1} }$

$\implies x=\sqrt{\dfrac{(\sqrt{10} +1)^{2} }{(\sqrt{10} )^{2} -1^{2} } }$

$\implies x=\sqrt{\dfrac{(\sqrt{10} +1)^{2} }{9} }$

$\implies x=\dfrac{\sqrt{10} +1}{3}$

How do we derive a quadratic equation having $x=\dfrac{\sqrt{10} +1}{3}$ as a solution? We can use squaring method to get the quadratic equation. So, the value of $x$ satisfies the following.

$\implies 3x-1=\sqrt{10}$

$\implies (3x-1)^{2} =(\sqrt{10} )^{2}$

$\implies 9x^{2}-6x+1=10$

$\implies 9x^{2}-6x-9=0$

Dividing both sides by $x$, we get the following.

$\implies 9x-6-\dfrac{9}{x} =0$

$\implies 9x-\dfrac{9}{x} =6$

Let's divide both sides by 9, and here is the answer.

$\therefore x-\dfrac{1}{x} =\dfrac{2}{3}$

This is the required answer.

Answer 2

Given that,

$\qquad \bigstar \:\:\underline {\boxed {\pmb{\red{ \sf \:x\:=\: \sqrt{\dfrac{\sqrt{10}+1 }{\sqrt{10}-1 } } \:\:}}}}\:$

Need To Find : Value of x - 1 /x ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

$\qquad \dashrightarrow \sf \:x\:=\: \sqrt{\dfrac{\sqrt{10}+1 }{\sqrt{10}-1 } } \:$

$\qquad \bigstar \:\:\underline {\pmb{\purple {\sf By \:Rationalizing \:the \:value \:of \:x \:\:\:}}}$

$\qquad \dashrightarrow \sf \:x\:=\: \sqrt{\dfrac{\sqrt{10}+1 }{\sqrt{10}-1 } } \:$

$\qquad \dashrightarrow \sf \:x\:=\: \sqrt{\dfrac{\sqrt{10}+1 }{\sqrt{10}-1 } }\:\;\times \:\sqrt{\dfrac{\sqrt{10}+1 }{\sqrt{10}+ 1 } } \:$

$\qquad \dashrightarrow \sf \:x\:=\: \sqrt{\dfrac{\big( \sqrt{10}+1 \big)^2 }{\big( \sqrt{10}\big)^2 -\big( 1 \big)^2 } }\:\; \:$

$\qquad \dashrightarrow \sf \:x\:=\: \sqrt{\dfrac{\big( \sqrt{10}+1 \big)^2 }{\big( 10 \big) -\big( 1 \big) } }\:\; \:$

$\qquad \dashrightarrow \sf \:x\:=\: \sqrt{\dfrac{\big( \sqrt{10}+1 \big)^2 }{ 9 } }\:\; \:$

$\qquad \dashrightarrow \sf \:x\:=\: \dfrac{ \sqrt{10}+1 }{ 3 } \:\; \:$

$\qquad \dashrightarrow \sf \:3x\:-\:1 \:=\: \sqrt{10}\:\; \:$

$\qquad \bigstar \:\:\underline {\pmb{\purple {\sf By \: Squaring \:both \:side \:\:\:}}}$

$\qquad \dashrightarrow \sf \big(\:3x\:-\:1 \:\big)^2\:=\: \big( \sqrt{10}\:\big)^2\; \:$

$\qquad \dashrightarrow \sf \:\big(3x\big)^2\:-\:6x \:+\:1 \:=\: 10\:\; \:$

$\qquad \dashrightarrow \sf \:9x^2\:-\:6x \:+\:1 \:=\: 10\:\; \:$

$\qquad \dashrightarrow \sf \:9x^2\:-\:6x \:- \:9 \:=\: 0\:\; \:$

$\qquad \dashrightarrow \sf \:\dfrac{9x^2}{x}\:-\:\dfrac{6x}{x} \:- \:\dfrac{9}{x} \:=\: \dfrac{0}{x}\:\; \:$

$\qquad \dashrightarrow \sf \:9x\:-\:6 \:- \dfrac{\:9}{x} \:=\: 0\:\; \:$

$\qquad \dashrightarrow \sf \:9x\: \:- \dfrac{\:9}{x} \:=\: 6\:\; \:$

$\qquad \dashrightarrow \sf \dfrac{\:9x\: \:- \dfrac{\:9}{x}}{9} \:=\: \dfrac{6}{9}\:\; \:$

$\qquad \dashrightarrow \sf \:x\: \:- \dfrac{\:1}{x} \:=\: \dfrac{2}{3}\:\; \:$

$\qquad \dashrightarrow \underline {\boxed {\purple {\pmb{\frak{ \:x\: \:- \dfrac{\:1}{x} \:=\: \dfrac{2}{3}\:\; }}}}}\:\:\bigstar$

$\therefore \underline {\sf Hence, \:\:The \:Value \:of \:x - 1/x \:is\:\pmb{\bf{ 2/3}}\:.}$