Question

factorisation .6√3x^2+7x=√3​

Answer 1

$6 \sqrt{3} {x}^{2} + 7x - \sqrt{3} = 0 \\ 6 \sqrt{3} {x}^{2} - 2x + 9x - \sqrt{3} = 0 \\ 2x(3 \sqrt{3} x - 1) + \sqrt{3} (3 \sqrt{3} x - 1) = 0 \\ (2x + \sqrt{3} )(3 \sqrt{3} x - 1) = 0$

Answer 2

Answer:

$\bold\red{(2x + \sqrt{3} )(3 \sqrt{3}x - 1) = 0}$

Step-by-step explanation:

Given,

$6 \sqrt{3} {x}^{2} + 7x = \sqrt{3}$

We may write it as

$6 \sqrt{3} {x}^{2} + 7x - \sqrt{3} = 0$

Let's check the product of,

coefficient of ${x}^{2}$ and constant term,

That is,

$= 6 \sqrt{3} \times (- \sqrt{3} )\\ \\ = - 18$

Now ,

we have to do middle term splitting,

$= > 6 \sqrt{3} {x}^{2} + 9x - 2x - \sqrt{3}$

Taking out common terms,

we get,

$= > 3 \sqrt{3}x (2x + \sqrt{3} ) - 1(2x + \sqrt{3} ) = 0 \\ \\ = > \bold{(2x + \sqrt{3} )(3 \sqrt{3}x - 1) = 0}$