How would one rationalize $\frac6{(1-\sqrt3)}$?
My answer is $\frac{6+6\sqrt3}{-2}$ ? Is this correct? Not sure its in the simplified form..? Simple pre-calc but apparently I am stuck. Thank you.
How would one rationalize $\frac6{(1-\sqrt3)}$?
My answer is $\frac{6+6\sqrt3}{-2}$ ? Is this correct? Not sure its in the simplified form..? Simple pre-calc but apparently I am stuck. Thank you.
Do not omit braces
$$\frac{6+6\sqrt3}{-2}$$ is correct but observe that you can still cancel $2$ from the numerator and the denominator.
The trick to solve the question is to multiply $1+\sqrt3$ (i.e. the conjugate to both the top and bottom) and use the formula of $(a-b)(a+b)=a^2-b^2$.