Question

please solve this only question no:3​

Answer 1

hey mate

here is your answer

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

$= {y}^{2} + 2 \sqrt{3} y - 9 \\ = {y}^{2} + 3 \sqrt{3} y - \sqrt{3} y - 9 \\ = y(y + \sqrt{3} ) - 3 \sqrt{3} (y + \sqrt{3} ) \\ = (y + 3 \sqrt{3} )(y - 3) \\ y = 3 \: and \: - 3 \sqrt{3}$

hope it helps u

ii)

$= {x}^{2} - 3 \sqrt{3} x + 6 \\ = {x}^{2} + 2 \sqrt{3} x + \sqrt{3} x + 6 \\ = x(x + \sqrt{3} ) + 2 \sqrt{3} (x + \sqrt{3} ) \\ = (x + 2)(x + 2 \sqrt{3} ) \\ x = - 2 \: and \: - 2 \sqrt{3}$

iii

y^2+3√3y-√3y-9

y(y-3)-3√3(y-3)

y= 3 & -3√3