Question

question no ten Plzz with solutions​

Answer 1

☯ Given :-

$\sf\ \bigg(x+\dfrac{1}{x}\bigg)^2 =4+\dfrac{3}{2}\bigg(x-\dfrac{1}{x}\bigg)$

☯To Find out :-

• We have to find the value of 'x'

$\underline{\sf{\dag\ \ Formula \ used :-\ \ }}$

$\bullet\sf \ \ (a+b)^2= (a-b)^2+4ab\\ \\ \bullet\sf (a-b)^2= a^2+b^2-2ab$

Solution :-

$:\implies\sf Suppose \ \bigg(x-\dfrac{1}{x}\bigg)= a \\ \\ :\implies\sf\ then \ \bigg(x+\dfrac{1}{x}\bigg)^2= \bigg(x-\dfrac{1}{x}\bigg)^2+4\times \cancel{x}\times \dfrac{1}{\cancel{x}}\\ \\ :\implies\sf \bigg(x+\dfrac{1}{x}\bigg)^2=a^2+4 ---- eq.\ (i)$

☯$\underline{\boldsymbol{\ According\ to\ Question:}}$

$\dashrightarrow\sf \bigg(x+\dfrac{1}{x}\bigg)^2 = 4+\dfrac{3}{2}\bigg(x-\dfrac{1}{x}\bigg)\\ \\ \\ \dashrightarrow\sf a^2+4=4+\dfrac{3}{2}a\ \ \ \ \ \therefore\Big[ From \ eq.\ (i) \Big]\\ \\ \\ \dashrightarrow\sf a^2-\dfrac{3a}{2}= 4-4 \\ \\ \\ \dashrightarrow\sf a\bigg(a-\dfrac{3}{2}\bigg)= 0 \ \ \ \ \therefore \bigg[ taking \ " a" \ as \ common\bigg] \\ \\ \\ \dashrightarrow\sf a=0 \ \ \ ;\ \ a-\dfrac{3}{2}=0\\ \\ \\ \dashrightarrow{\purple{\sf a=0 \ \ ;\ \ a=\dfrac{3}{2}}}$

$\underline{\bigstar{\sf The \ value \ of \ a \ is \ either \ 0\ \ or\ \ 3/2}}$

★Now we have to find the value of x

☯ Now substitute the value of a=0

$:\implies\sf \bigg(x-\dfrac{1}{x}\bigg)^2= 0\\ \\ \\ :\implies\sf \dfrac{(x^2-1)}{x}=0\\ \\ \\ :\implies\sf x^2-1=0\\ \\ \\ :\implies\sf x^2= 1 \\ \\ \\ :\implies {\red{\sf\ x= \pm\ 1 }}\ \ \ \ \bigg[ \pm\ 1= 1 , \ and \ (-1)\bigg]$

☯ Then substitute the value of a=3/2

$:\implies\sf \bigg(x-\dfrac{1}{x}\bigg)= \dfrac{3}{2}\\ \\ \\ :\implies\sf \dfrac{x^2-1}{x}= \dfrac{3}{2}\\ \\ \\ :\implies\sf 2(x^2-1)=3x\ \ \ \ \therefore\bigg[\ By \ Cross \ multiplication \bigg]\\ \\ \\ :\implies\sf 2x^2-3x-2=0\\ \\ \\ :\implies\sf\ 2x^2-(4-1)x-2=0\ \ \ \ \therefore \bigg( splitting\ middle \ term \bigg)\\ \\ \\ :\implies\sf 2x^2-4x+x-2=0\\ \\ \\. :\implies\sf 2x(x-2)+1(x-2)=0\\ \\ \\ :\implies\sf (x-2)(2x+1)=0\\ \\ \\ :\implies{\red{\sf x=2 \ \ ; \ \ x= \dfrac{-1}{2}}}$

$\therefore{\underline{\sf Hence , \ the \ value \ of \ x\ =\ 1 , \ (-1)\ , \ 2,\ and \ (-1/2)}}$

Answer 2

$</p><p>\green{\underline \bold{Given :}}</p><p></p><p>$

$\bigg(x+\dfrac{1}{x}\bigg)^2 =4+\dfrac{3}{2}\bigg(x-\dfrac{1}{x}\bigg)$

$\red{\underline \bold{To \: Find:}}</p><p></p><p>$

• we will find the value of x

$\blue{\boxed{ \bold{Some \: related \: formula \: will \: be \: used}}}$

$\bullet\sf \ \ (a+b)^2= (a-b)^2+4ab$

$\bullet\sf (a-b)^2= a^2+b^2-2ab$

$</p><p>\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}</p><p></p><p>$

$\implies\sf Suppose \ \bigg(x-\dfrac{1}{x}\bigg)= a$

$\sf\implies\ then \ \bigg(x+\dfrac{1}{x}\bigg)^2= \bigg(x-\dfrac{1}{x}\bigg)^2+4\times \cancel{x}\times \dfrac{1}{\cancel{x}}$

$\sf \implies\bigg(x+\dfrac{1}{x}\bigg)^2=a^2+4 ---- eq.\ (i)$

${\boldsymbol{\ According\ to\ Question:}}$

$\dashrightarrow\sf \bigg(x+\dfrac{1}{x}\bigg)^2 = 4+\dfrac{3}{2}\bigg(x-\dfrac{1}{x}\bigg)$

$\dashrightarrow\sf a^2+4=4+\dfrac{3}{2}a$

$therefore\Big[ From \ eq.\ (i) \Big]$

$\dashrightarrow\sf a^2-\dfrac{3a}{2}= 4-4$

$\dashrightarrow\sf a\bigg(a-\dfrac{3}{2}\bigg)= 0$

$\therefore \bigg[ taking \ " a" \ as \ common\bigg]$

$\dashrightarrow\sf a=0$

$a-\dfrac{3}{2}=0$

$\dashrightarrow{\purple{\sf a=0 \ \ ;\ \ a=\dfrac{3}{2}}}$

$\underline{{\sf The \ value \ of \ a \ is \ either \ 0\ \ or\ \ 3/2}}$

• Now we will find the value of x

• Substitute the value of a = 0

$\implies\sf \bigg(x-\dfrac{1}{x}\bigg)^2= 0$

$\implies\sf \dfrac{(x^2-1)}{x}=0$

$\implies\sf x^2-1=0$

$\implies\sf x^2= 1$

$\implies {\red{\sf\ x= \pm\ 1 }}$

$\bigg[ \pm\ 1= 1 , \ and \ (-1)\bigg]$

$\tt: \implies Now \: substitute \: the \: value \: of \: a \: = \frac{3}{2}$

$\implies\sf \bigg(x-\dfrac{1}{x}\bigg)= \dfrac{3}{2}$

$\implies\sf \dfrac{x^2-1}{x}= \dfrac{3}{2}$

$\implies\sf 2(x^2-1)=3x$

$\therefore\bigg[\ By \ Cross \ multiplication \bigg]$

$\implies\sf 2x^2-3x-2=0$

$\implies\sf\ 2x^2-(4-1)x-2=0$

$\therefore \bigg( splitting\ middle \ term \bigg)$

$\implies\sf 2x^2-4x+x-2=0$

$\implies\sf 2x(x-2)+1(x-2)=0$

$\implies\sf (x-2)(2x+1)=0$

$\implies{\red{\sf x=2 \ \ ; \ \ x= \dfrac{-1}{2}}}$

${\underline {\sf Hence , \ the \ value \ of \ x\ =\ 1 , \ (-1)\ , \ 2,\ and \ (-1/2)}}$