Question 356972
1. {{{sqrt(3)*(2sqrt(15)-3sqrt(4))}}}
Since {{{sqrt(4) = 2}}}, we can start by substituting:
{{{sqrt(3)*(2sqrt(15)-3(2))}}}
or
{{{sqrt(3)*(2sqrt(15)-6)}}}
Now we can multiply, using the Distributive Property:
{{{sqrt(3)*2sqrt(15)-sqrt(3)*6}}}
In the first product we use the property of radicals, {{{root(a, p)*root(a, q) = root(a, p*q)}}}, to multiply the two square roots. In the second factor we just use the Commutative Property to move the 6 in front:
{{{2sqrt(3*15)-6sqrt(3)}}}
{{{2sqrt(45)-6sqrt(3)}}}
We cannot subtract these two terms because they are not like terms. But we can simplify the first square root. (It has a perfect square factor: 9)
{{{2sqrt(9*5)-6sqrt(3)}}}
Using the same property of radicals as above (only "in reverse") we can split these two square roots:
{{{2sqrt(9)*sqrt(5)-6sqrt(3)}}}
Since {{{sqrt(9) = 3}}}:
{{{2(3)*sqrt(5)-6sqrt(3)}}}
or
{{{6sqrt(5)-6sqrt(3)}}}<br>
2. {{{(3+sqrt(11))*(3-sqrt(11))}}}
We can use FOIL to multiply this. Or we can take advantage of the difference of squares pattern, {{{(a+b)(a-b) = a^2 - b^2}}}, with "a" being 3 and "b" being {{{sqrt(11)}}}:
{{{(3)^2 - (sqrt(11))^2}}}
9 - 11
-2<br>
3. {{{(sqrt(5)-sqrt(3))^2}}}
For this we can use FOIL. Or we can use a pattern, {{{(a-b)^2 = a^2 -2ab + b^2}}} with "a" being {{{sqrt(5)}}} and "b" being {{{sqrt(3)}}}:
{{{(sqrt(5))^2 - 2(sqrt(5))(sqrt(3)) + (sqrt(3))^2}}}
{{{5 -2sqrt(15) + 3}}}
{{{8 - 2sqrt(15)}}}<br>
Note: Even if you use FOIL on #2 and #3, you still end up with the same answers as we did using patterns.