Question

2x^2-7x+3=0 solve by completing ​

Answer 1

Hey your answer is in the attachment

Hope it helps.

Answer 2

$\huge\boxed{x = 3}$$\LARGE{ \: \: or \: \: }$$\huge\boxed{x = \frac{1}{2} }$

Answer:

Method used : Completing the square

We have,

$\Large{ 2 {x}^{2} - 7x + 3 = 0}$

$\Large{ \Rightarrow {x}^{2} - \frac{7}{2} x + \frac{3}{2} = 0}$———————[dividing both the sides with 2]

$\Large{ \Rightarrow {x}^{2} - \frac{7}{2} x = - \frac{3}{2} }$

$\Large{ \Rightarrow {x}^{2} - 2(x) \left( \frac{7}{4} \right) = - \frac{3}{2} }$

$\Large{ \Rightarrow {x}^{2} - 2(x) \left( \frac{7}{4} \right) + { \left( \frac{7}{4} \right) }^{2} = - \frac{3}{2} + { \left( \frac{7}{4} \right) }^{2} }$——-————-[adding${ \left( \frac{7}{4} \right) }^{2}$to both the sides]

$\Large{ \Rightarrow { \left( x - \frac{7}{4} \right) }^{2} = - \frac{3}{2} + { \left( \frac{7}{4} \right) }^{2} }$—-——————( since ${(a - b)}^{2} = {a}^{2} - 2ab + {b}^{2}$ )

$\Large{ \Rightarrow { \left( x - \frac{7}{4} \right) }^{2} = - \frac{3}{2} + \frac{49}{16} }$

$\Large{ \Rightarrow { \left( x - \frac{7}{4} \right) }^{2} = \frac{49 - 24}{16} }$

$\Large{ \Rightarrow { \left( x - \frac{7}{4} \right) }^{2} = \frac{25}{16} }$

$\Large{ \Rightarrow { \left( x - \frac{7}{4} \right) }^{2} = { \left( \pm \frac{5}{4} \right) }^{2} }$

$\Large{ \Rightarrow x - \frac{7}{4} = \pm \frac{5}{4} }$———————[on square-rooting both the sides]

$\Large{ \Rightarrow x = \frac{7}{4} \pm \frac{5}{4} }$

$\Large{ \Rightarrow x = \frac{7 \pm 5}{4} }$

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$\Large{ \therefore x = \frac{7 + 5}{4} \implies x = \frac{12}{4} = \frac{ \cancel{12} \: {}^{3} }{ \cancel{4} _1} = 3}$

OR

$\Large{ \therefore x = \frac{7 - 5}{4} \implies x = \frac{2}{4} = \frac{ \cancel{2} \: {}^{1} }{ \cancel{4} _2} = \frac{1}{2} }$

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