Question 400987
I think by "index 3 sqrt 3" you mean {{{root(3, 3)}}}. If this is correct, then it is called the cube root of 3. It is not a square root of any kind.<br>
Your expression:
{{{(x+sqrt(3))(x-root(3, 3))}}} 
is correct. But we need an equation, not an expression. The equation is simply
{{{(x+sqrt(3))(x-root(3, 3)) = 0}}} 
Now we just need it in {{{ax^2+bx+c=0}}} form. For this we just multiply out the left side using FOIL:
{{{x^2-x*root(3, 3)+(sqrt(3))x-sqrt(3)*root(3, 3)=0}}}
What follows will make more sense if we rewrite the equation as additions. (Also, the {{{ax^2+bx+c=0}}} form is written as additions.):
{{{x^2+ (-x*root(3, 3)) + (sqrt(3))x+ (-sqrt(3)*root(3, 3))=0}}}
Factoring out x from the middle two terms we get:
{{{x^2 + (-root(3, 3)+ sqrt(3))x + (-sqrt(3)*root(3, 3))=0}}}
If you have not yet learned about fractional exponents then the equation above is your answer with...
"a" being 1
"b" being {{{(-root(3, 3)+ sqrt(3))}}}
and "c" being {{{(-sqrt(3)*root(3, 3))}}}<br>
If you do know about fractional exponents then we can simplify {{{(sqrt(3)*root(3, 3))}}}. Rewriting this with fractional exponents we get:
{{{3^(1/2)*3^(1/3)}}}
The rule for exponents when multiplying is to add the exponents. These exponents are fractions and to add fractions we need a common denominator:
{{{3^(3/6)*3^(2/6)}}}
Now we can multiply:
{{{3^((3/6)+(2/6))}}}
{{{3^(5/6)}}}
We can now write this back in radical form:
{{{root(6, 3^5)}}}
{{{3^5 = 243}}} so this becomes:
{{{root(6, 243)}}}
This makes our full, simplified equation:
{{{x^2 + (-root(3, 3)+ sqrt(3))x + (-root(6, 243))=0}}}
and ...
"a" being 1
"b" being {{{(-root(3, 3)+ sqrt(3))}}}
and "c" being {{{(-root(6, 243))}}}<br>
(Note: The "b", {{{(-root(3, 3)+ sqrt(3))}}}, cannot be simplified, even if you use fractional exponents. They are not like terms and cannot be transformed into like terms. So we can never add them.)