Question

Factorise

$4 \sqrt{3 {x}^{2} } + 5x - 2 \sqrt{3}$

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Answer 1

Correct QuestioN :

Factorise $\tt \: \: \: \: \: 4 \sqrt{3} {x}^{2} + 5x - 2 \sqrt{3} .$

SolutioN :

We have, Expression.

By Splitting the Middle Terms.

→ 4√3x² + 5x - 2√3.

→ 4√3x² + 8x - 3x - 2√3.

→ 4x(√3x + 2 ) - √3( √3x + 2 )

→ ( 4x - √3 )( √3 + 2 )

Either,

→ 4x - √3 = 0.

→ x = √3/ 4.

Or,

→ √3x + 2 = 0.

→ x = - 2 / √3.

$\tt \dagger \: \: \: \: \: 4 \sqrt{3} {x}^{2} + 5x - 2 \sqrt{3} .$

$\tt \dagger \: \: \: \: \: a {x}^{2} + bx + c.$

Where as,

• a = 4√3.
• b = 5.
• c = - 2√3.

Let Try to solve by Quadratic Formula.

$\dagger \: \: \: \: \: \boxed{ \tt{x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} }}$

$\tt : \implies x = \dfrac{ - 5\pm \sqrt{ {(5)}^{2} - 4 \times 4 \sqrt{3} \times - 2 \sqrt{3} } }{2 \times 4 \sqrt{3} }$

$\tt : \implies x = \dfrac{ - 5\pm \sqrt{ 25 + 96 } }{ 8 \sqrt{3} }$

$\tt : \implies x = \dfrac{ - 5\pm \sqrt{ 121 } }{ 8 \sqrt{3} }$

$\tt : \implies x = \dfrac{ - 5\pm 11 }{ 8 \sqrt{3} }$

Either,

$\tt : \implies x = \dfrac{ - 5 + 11 }{ 8 \sqrt{3} }$

$\tt : \implies x = \dfrac{ 6}{ 8 \sqrt{3} }$

$\tt : \implies x = \dfrac{3 }{ 4 \sqrt{3} }$

$\tt : \implies x = \dfrac{3 }{ 4 \sqrt{3} } \times \dfrac{ \sqrt{3} }{ \sqrt{3} }$

$\tt : \implies x = \dfrac{3 \sqrt{3} }{ 12}$

$\tt : \implies x = \dfrac{\sqrt{3} }{ 4 }$

Or,

$\tt : \implies x = \dfrac{ - 5 - 11 }{ 8 \sqrt{3} }$

$\tt : \implies x = \dfrac{ - 16 }{ 8 \sqrt{3} }$

$\tt : \implies x = \dfrac{ - 2}{ \sqrt{3} }$

$\tt : \implies x = \dfrac{ - 2}{ \sqrt{3} } \times \dfrac{ \sqrt{3} }{ \sqrt{3} }$

$\tt : \implies x = \dfrac{ - 2 \sqrt{3} }{ 3 }$

Therefore, the value of x is - 2 √3 / 3 and 3 / 4√3.