Question

Factorise √3x²+ x - 10 √3

Answer 1

Solution :

We have,

$\tt \dagger \: \: \: \: \: \: \sqrt{3} {x}^{2} + x - 10 \sqrt{3}.$

Let Find there Zeros.

• Splitting the middle term.
• Quadratic Formula.

By Quadratic Formula.

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

Where as,

• a = √3.
• b = 1.
• c = -10√3.

$\tt \longmapsto x = \dfrac{ - 1 \pm \sqrt{ {1}^{2} - 4 \sqrt{3} \times - 10 \sqrt{3} }}{2 \sqrt{3} }$

$\tt \longmapsto x = \dfrac{ - 1 \pm \sqrt{120+ 1 } }{2 \sqrt{3} }$

$\tt \longmapsto x = \dfrac{ - 1 \pm \sqrt{121 } }{2 \sqrt{3} }$

$\tt \longmapsto x = \dfrac{ - 1 \pm \sqrt{11 \times 11 } }{2 \sqrt{3} }$

$\tt \longmapsto x = \dfrac{ - 1 \pm 11 }{2 \sqrt{3} }$

Either,

$\tt \longmapsto x = \dfrac{ - 1 + 11 }{2 \sqrt{3} }$

$\tt \longmapsto x = \dfrac{ 10 }{2 \sqrt{3} }$

$\tt \longmapsto x = \dfrac{ 5 }{ \sqrt{3} }$

Or,

$\tt \longmapsto x = \dfrac{ - 1 - 11 }{2 \sqrt{3} }$

$\tt \longmapsto x = \dfrac{ -12 }{2 \sqrt{3} }$

$\tt \longmapsto x = \dfrac{ - 6}{ \sqrt{3} }$

Therefore, the value of x is -6 /√3 and 5/√3.