Question

6x²+9√3 x+7 factories​

Answer 1

$\huge \bold\dag \mathfrak\color{orange}A \color{red}n \color{green}s \color{blue}w \color {purple}e \color {lime}r$

We use quadratic formula method for factorisation.

a = 6 ; b = 9√3 ; c = 7

D = b² - 4ac

= (9√3)² - 4 x 6 x 7

= 81 x 3 - 168

= 243 - 168

= 75 > 0

$\therefore \: it \: has \: 2 \: distinct \: real \: roots$

$\large{x \: = \frac{ - b± \sqrt{ D } }{2a} }$

$\large{x \: = \frac{ - 9 \sqrt{3} \: ± \: \sqrt{75} }{2 \times 6} }$

$\:\:\:\:\:\:\:\:\:(√75 = √25 \times √3 = 5√3)$

$\large{x \: = \frac{ -9 \sqrt{3} \: ± \: 5 \sqrt{3} }{12}}$

$\large{x = \frac{9 \sqrt{3} + 5 \sqrt{3} }{12} \: (or) \: x = \frac{9 \sqrt{3} - 5 \sqrt{3} }{12}}$

$\large{x = \frac{14 \sqrt{3} }{12} \: (or) \: x = \frac{4 \sqrt{3} }{12}}$

$\large {x = \frac{7 \sqrt{3} }{6} \: (or) \: x = \frac{2 \sqrt{3} }{6}}$

Answer 2

Answer:

I don't know your answer sorry