Question

Solve question no. 11 by completing the square method.

Answer 1

Question :-

Solve the equation ( 2/x² ) - ( 5/x ) + 2 = 0 by completing square method.

Solution :-

$\sf \dfrac{2}{ {x}^{2} } - \dfrac{5}{x} + 2 = 0$

Taking LCM

$\implies \sf \dfrac{2 - 5(x) + 2( {x}^{2} )}{ {x}^{2} }= 0$

$\implies \sf \dfrac{2 - 5x+ 2 {x}^{2} }{ {x}^{2} }= 0$

$\implies \sf 2 - 5x+ 2 {x}^{2} = 0( {x}^{2})$

$\implies \sf 2 - 5x+ 2 {x}^{2} = 0$

$\implies \sf 2 {x}^{2} - 5x+ 2 = 0$

$\implies \sf 2 {x}^{2} - 5x = - 2$

Dividing throuout by '2'

$\implies \sf \dfrac{2 {x}^{2} }{2} - \dfrac{5x}{2} = - \dfrac{2}{2}$

$\implies \sf {x}^{2} - \dfrac{5x}{2} = - 1$

$\implies \sf {x}^{2} - 2(x) \bigg( \dfrac{5}{4} \bigg) = - 1$

Adding ( 5/4 )² on both sides

$\implies \sf {x}^{2} - 2(x) \bigg( \dfrac{5}{4} \bigg) + \bigg( \dfrac{5}{4} \bigg)^{2} = - 1 + \bigg( \dfrac{5}{4} \bigg)^{2}$

$\implies \sf \bigg(x - \dfrac{5}{4} \bigg)^{2} = - 1 + \dfrac{ {5}^{2} }{ {4}^{2} }$

[ Because a² - 2ab + b² = (a - b)² ]

$\implies \sf \bigg(x - \dfrac{5}{4} \bigg)^{2} = - 1 + \dfrac{25}{ 16 }$

$\implies \sf \bigg(x - \dfrac{5}{4} \bigg)^{2} = \dfrac{ - 16 + 25}{ 16 }$

$\implies \sf \bigg(x - \dfrac{5}{4} \bigg)^{2} = \dfrac{9}{ 16 }$

Taking square root on both sides

$\implies \sf x - \dfrac{5}{4} = \pm\sqrt{\dfrac{9}{ 16 }}$

$\implies \sf x - \dfrac{5}{4} = \pm \dfrac{3}{4}$

$\implies \sf x - \dfrac{5}{4} = \dfrac{3}{4} \ \ or \ \ x - \dfrac{5}{4} = - \dfrac{3}{4}$

$\implies \sf x = \dfrac{3}{4} + \dfrac{5}{4} \ \ or \ \ x = - \dfrac{3}{4} + \dfrac{5}{4}$

$\implies \sf x = \dfrac{3 + 5}{4} \ \ or \ \ x = \dfrac{ - 3 + 5}{4}$

$\implies \sf x = \dfrac{8}{4} \ \ or \ \ x = \dfrac{2}{4}$

$\implies \sf x = 2 \ \ or \ \ x = \dfrac{1}{2}$

Hence, 2 and 1/2 are the roots of the equation.

Answer 2

Question :-

Solve this quadratic by completing the square method .

$\sf{\frac{2}{ {x}^{2} } - \frac{5}{x} + 2 = 0}$

Answer :-

→ Roots are 2 and 1/2

Step - by - step explanation :-

Given equation is ,

$\rightarrow \: \sf{\red{\frac{2}{ {x}^{2} } - \frac{5}{x} + 2 = 0}} \\ \: \\ \rightarrow \sf{ \frac{2 - 5x + 2 {x}^{2} }{ {x}^{2} } = 0} \\ \\ \rightarrow \sf{\green{ 2 {x}^{2} - 5x + 2 = 0}}$

Now ,we using completing the square method ,

$\rightarrow \sf{\red{2 {x}^{2} - 5x + 2 = 0}} \\ \\ \sf{divided \: by \: \: 2} \\ \\ \rightarrow \sf{\green{\frac{2 {x}^{2} }{2} - }\red{\frac{5x}{2} + \frac{2}{2}} = \frac{0}{2} } \\ \\ \rightarrow \sf{ {x}^{2} - \frac{5x}{2} = - 1} \\ \\ \sf{\green{ to \: making \: it \: a \: perfect \: square \:} } \\ \sf{adding \: { \bigg(\frac{5}{4} \bigg)}^{2} \: on \: both \: sides \: } \\ \\ \rightarrow \red{ \sf{ {x}^{2} + { \big( \frac{5}{4} \big) }^{2} }+ \green{2x \times \frac{5}{4} } = - 1 + { \bigg( \frac{5}{4} \bigg) }^{2} } \\ \\ \rightarrow \red{\sf{ { \bigg(x - \frac{5}{4} \bigg)}^{2} } = \green{ \frac{ - 16 + 25}{16} }} \\ \\ \rightarrow \: \pink{ \sf{ { \bigg(x - \frac{5}{4} \bigg)}^{2} \: } }= \red{ \frac{9}{16} } \\ \\ \blue{ \sf{taking \: \sqrt{ \: \: } \: on \: both \: sides \: } }\\ \\ \rightarrow \red{\sf{ x - \frac{5}{4} = - \frac{3}{4} }\green{ \: or \: + \frac{3}{4} }} \\ \\ \star \: \sf{case \: (1)} \\ \\ \implies \red{\sf{x - \frac{5}{4}} = \green{ - \frac{3}{4} } }\\ \\ \implies \sf{x = - \frac{ 3}{4} + \frac{5}{4} } \\ \\ \implies \sf{ x \: = \frac{ 2}{4} } \\ \\ \implies \: \boxed{ \sf{ x = \frac{1}{2} }} \\ \\ \star \sf{case \: (2)} \\ \\ \implies \red{ \sf{ x - \frac{5}{4} = \frac{3}{4} } }\\ \\ \implies \sf{ x \: = \frac{8}{4} } \\ \\ \implies \boxed{\sf{x \: = 2}}$

2 and 1/2 are the roots of this equation.

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