Question

$x {}^{2} - \sqrt{3}x - 6 = {?}^{}$
please solve the problem ​

Answer 1

Answer:

(x+√3)(x-2√3)

Step-by-step explanation:

x²-√3x-6

=x²-2√3x+√3x-6

=x(x-2√3)+√3(x-2√3)

=(x+√3)(x-2√3)

Pls mark as the brainliest :)

Answer 2

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

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

$\rightarrow \: x(x - 2 \sqrt{3} ) - \sqrt{3} (x - 2 \sqrt{3}) = 0$

$\rightarrow \: (x - 2 \sqrt{3} ) (x - \sqrt{3}) = 0$

So the roots are either

$\boxed{2 \sqrt{3} }$

or

$\boxed{ \sqrt{3} }$