Question

Solve by factorization $\sqrt{3}{x}^2+10x+7\sqrt{3}=0$

• A. $x=-\sqrt{3},-\frac{7}{\sqrt{3}}$
• B. $x=-\sqrt{3},\frac{7}{\sqrt{3}}$
• C. $x=2,\frac{1}{2}$
• D. ${+}_{-}3$

Answer 1

The correct option is A $x=-\sqrt{3},-\frac{7}{\sqrt{3}}$

$\sqrt{3}{x}^2+10x+7\sqrt{3}=0$

$a=\sqrt{3},b=10,c=7\sqrt{3}$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-10\pm \sqrt{{10}^2-4\times 7\sqrt{3}\times \sqrt{3}}}{2\sqrt{3}}=\frac{-10\pm \sqrt{16}}{2\sqrt{3}}$

$x=\frac{-10\pm 4}{2\sqrt{3}}$

$x=\frac{-10+4}{2\sqrt{3}},x=\frac{-10-4}{2\sqrt{3}}$

$\therefore x=\frac{-6}{2\sqrt{3}}=-\sqrt{3};\therefore x=-\frac{7}{\sqrt{3}}$

$\therefore x=-\sqrt{3},-\frac{7}{\sqrt{3}}$