Question

Solve by using the method of completing the square :-

$1) \: \: \sqrt{2} x {}^{2} - 3x - 2 \sqrt{2} = 0 \\ \\ 2) \: \: \sqrt{3} x {}^{2} + 10x + 7 \sqrt{3}$

Answer 1

Hey there !!


➡ Question 1 :-)

The given quadratic equation :-

$\: \: \sqrt{2} x {}^{2} - 3x - 2 \sqrt{2} = 0$

[ Multiplying both side by $\sqrt{2}$ ] .

=> 2x² - 3$\sqrt{2}$x - 4 = 0.

=> 2x² - 3$\sqrt{2}$x = 4 .


$[ \: adding {( \frac{3}{2}) }^{2} on \: both \: side \: ]. \\ \\ = > {( \sqrt{2} x)}^{2} - 2 \times \sqrt{2} x \times \frac{3}{2} + {( \frac{3}{2}) }^{2} = 4 + {( \frac{3}{2}) }^{2} . \\ \\ = > {( \sqrt{2}x - \frac{3}{2} )}^{2} = 4 + \frac{9}{4} . \\ \\ = > {( \sqrt{2}x - \frac{3}{2} )}^{2} = \frac{16 + 9}{4} . \\ \\ = > {( \sqrt{2}x - \frac{3}{2} )}^{2} = {( \frac{5}{2}) }^{2} . \\ \\ ( \: taking \: square \: root \: on \: both \: side, \: we \: get \: ) \\ \\ = > \sqrt{2} x - \frac{3}{2} = ± \frac{5}{2} . \\ \\ = > \sqrt{2} x - \frac{3}{2} = \frac{5}{2} \: \: or \: \: \sqrt{2} x - \frac{3}{2} = - \frac{5}{2} . \\ \\ = > \sqrt{2} x = \frac{5}{2} + \frac{3}{2} \: \: or \: \: \sqrt{2} x = - \frac{5}{2} + \frac{3}{2} . \\ \\ = > \sqrt{2} x = \frac{8}{2} \: \: or \: \: \sqrt{2} x = \frac{ - 2}{2} . \\ \\ = > x = \frac{4}{ \sqrt{2} } \: \: or \: \: x = \frac{ - 1}{ \sqrt{2} } . \\ \\ = > x = \frac{ \sqrt{2} \times \sqrt{2} \times \sqrt{2} \times \cancel{\sqrt{2}} }{ \cancel{\sqrt{2}} } \: \: or \: \: x = \frac{ - 1}{ \sqrt{2} } . \\ \\ = > \boxed{ \bf \therefore x = 2 \sqrt{2} \: \: or \: \: x = \frac{ - 1}{ \sqrt{2} } .}$


➡ Question 2 :-)

The given quadratic equation:-

$\sqrt{3} x {}^{2} + 10x + 7 \sqrt{3} = 0 .$

[ Multiplying both side by $\sqrt{3}$ ] .

$= > 3 {x}^{2} + 10 \sqrt{3} x + 21 = 0. \\ \\ = > 3 {x}^{2} + 10 \sqrt{3} x = - 21. \\ \\ = > {( \sqrt{3}x) }^{2} + 2 \times ( \sqrt{3} x) \times 5 + {5}^{2} = - 21 + {5}^{2} . \\ \\ = > {( \sqrt{3}x + 5) }^{2} = - 21 + 25. \\ \\ = > {( \sqrt{3} + 5) }^{2} = {2}^{2} . \\ \\ (taking \: square \: root \: on \: both \: side \: we \: get \: ) \\ \\ = > \sqrt{3} x + 5 = 2 \: \: or \: \: \sqrt{3} x + 5 = 2. \\ \\ = > \sqrt{3} x = 2 - 5 \: \: or \: \: \sqrt{3} x = - 2 - 5. \\ \\ = > x = \frac{ - 3}{ \sqrt{3} } \: \: or \: \: x = \frac{ - 7}{ \sqrt{3} } . \\ \\ \boxed{ \bf \therefore x = - \sqrt{3} \: \: or \: \: x = \frac{ - 7}{ \sqrt{3} } .}$


✔✔ Hence, it is solved ✅ ✅.



THANKS



#BeBrainly.

Answer 2

( i ) :

Given equation : √2x^2 - 3x - 2√2 = 0

⇒ √2x^2 - 3x = 2√2

⇒ ( √2x^2 /√2 ) - ( 3x /√2 ) = ( 2√2 / √2 )

⇒ x^2 -  { 3x / √2 × √2 / √2 }  = 2

⇒ x^2 - ( 3√2x / 2 ) = 2

⇒ x^2-(3√2x/2)+( 3√2/4)^2=2+(3√2/4)^2

⇒ [ x - ( 3√2 / 4 ) ]^2 = 2 + 18 / 16

⇒ [ x - ( 3√2 / 4 ) ]^2 = 25 / 8

⇒ x - 3√2/4 = √(25/8) or - √(25/8)

⇒ x - 3√2/4 = 5/(2√2) or - 5/(2√2)

⇒ x - 3√2/4 = (5√2)/4 or - (5√2)/4

⇒ x = (5√2)/4 + (3√2)/4 or - (5√2)/4 + (3√2)/4

⇒ x = √2( 5+3)]4 or √2(-5+3)/4

⇒ x = 2√2 or - 1/√2

( ii )

Given equation : √3x^2 + 10x + 7√3 = 0

⇒ √3x^2 + 10x = - 7√3

⇒ ( √3x^2 / √3 ) + ( 10x / √3 ) = - ( 7√3 / √3 )

⇒ x^2 + (10x/√3) = - 7

⇒ x^2 + (10√3x)/3 = - 7

⇒ x^2 + (10√3x)/3 + [ (10√3)/6 ]^2 = -7+[ (10√3)/6 ]^2

⇒ [ x + (10√3/6) ]^2 = -7+[300/36]

⇒ [ x + (10√3/6) ]^2 = -7+25/3

⇒ [ x + (10√3/6) ]^2 = 4 / 3

⇒ x + (10√3/6) = √(4/3) or - √(4/3)

⇒ x + (10√3/6) = 2/√3 or - 2/√3

⇒ x + (5√3/3) = 2√3/3 or - 2√3 / 3

⇒ x = (2√3/3)-(5√3/3) or - (2√3/3)-(5√3/3)

⇒ x = √3(2-5)/3 or -√3(2+5)/3

⇒ x = -√3 or - 7√3/3