Question

answer it with a attachment....​

Answer 1

Answer:

given equation is a quadratic equation

$\sqrt{3} {x}^{2} + 14x - 5 \sqrt{3} = 0$

we know that

roots of quadratic equation

$= \frac{ - b ± \sqrt{ {b}^{2}-4ac } }{2a}$

now,

substitute in the above formula

$\frac{ - 14± \sqrt{ {14}^{2} - 4( \sqrt{3})( - 5 \sqrt{3} ) } }{2 \sqrt{3} }$

$= \frac{-14± \sqrt{196 + 60} }{2 \sqrt{3} }$

$= \frac{ - 14 ± \sqrt{256} }{2 \sqrt{3} }$

$= \frac{ - 14 + \sqrt{256} }{2} (or) \frac{ - 14 - \sqrt{256} }{2 \sqrt{3} }$

$= \frac{ - 14 + 16}{2 \sqrt{3} } (or) \frac{ - 14 - 16}{2 \sqrt{3} }$

$= \frac{2}{2 \sqrt{3} } (or) \frac{ - 30}{2 \sqrt{3} }$

$= \frac{1}{ \sqrt{3 } }( or) \frac{ - 30}{2 \sqrt{3} }$

rationalize the denominators of the roots

$= \frac{ \sqrt{3} }{3} (or) - 5 \sqrt{3}$

now,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = \frac{ \sqrt{3} }{3}$

(or)

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = - 5 \sqrt{3}$

Answer 2

Hi

Inga answer irruku

$\frac{ \sqrt{3} }{?3}$