Question 633226
Those equations may be scary because of the square roots, but they are just quadratic equations. All that the square roots can do is scare you and cause you to make mistakes in the calculations.
I'll show you several ways to solve them. You can chose your way to solve each one.
 
{{{4x^2-4sqrt(3)x+3=0}}}
SMART FACTORING:
{{{4x^2=(2x)^2}}} is a square.
I would say that {{{3=(-sqrt(3))^2}}} is another square.
Is {{{-4sqrt(3)x}}} twice the product of {{{2x}}} and {{{-sqrt(3)}}}?
Yes, {{{2*(2x)*(-sqrt(3))=-4sqrt(3)x}}} .
Then, {{{4x^2-4sqrt(3)x+3}}} is the square of a binomial.
{{{(2x-sqrt(3))^2=4x^2-4sqrt(3)x+3}}}
The equation can be written as
{{{(2x-sqrt(3))^2=0}}} --> {{{2x-sqrt(3)=0}}} --> {{{highlight(x=sqrt(3)/2)}}}
COMPLETING THE SQUARE:
{{{4x^2-4sqrt(3)x+3=0}}} --> {{{x^2-sqrt(3)x+3/4=0}}} --> {{{x^2-sqrt(3)x=-3/4}}}
Adding {{{3/4}}} (the square of -sqrt(3)/2) to both sides we form the square of {{{x-sqrt(3)/2}}} on the left side:
{{{x^2-sqrt(3)x=-3/4}}} --> {{{x^2-sqrt(3)x+3/4=-3/4+3/4}}} --> {{{(x-sqrt(3)/2)^2=0}}} --> {{{x-sqrt(3)/2=0}}} --> {{{highlight(x=sqrt(3)/2)}}}
USING THE QUADRATIC FORMULA:
For the generic quadratic equation {{{ax^2+bx+c=0}}} the solution(s) is/are
{{{x=(-b +- sqrt(b^2-4ac))/2a}}} (if what's under the square root is not negative)
In the case of {{{4x^2-4sqrt(3)x+3=0}}} , {{{a=4}}}, {{{b=-4sqrt(3)}}}, and {{{c=3}}}
Substituting:
{{{x=(-(-4sqrt(3)) +- sqrt((-4sqrt(3))^2-4*4*3))/(2*4)}}} --> {{{x=(4sqrt(3) +- sqrt((-4)^2(sqrt(3))^2-48))/8}}}
 
{{{x=(4sqrt(3) +- sqrt(16*3-48))/8}}} --> {{{x=(4sqrt(3) +- sqrt(48-48))/8}}}
 
{{{x=(4sqrt(3) +- sqrt(0))/8}}} --> {{{x=(4sqrt(3) +- sqrt(0))/8}}}
 
{{{x=(4sqrt(3) +- 0)/8}}} --> {{{x=4sqrt(3)/8}}} --> {{{highlight(x=sqrt(3)/2)}}}
 
{{{sqrt(2)x^2-3x+sqrt(2)=0}}}
USING THE QUADRATIC FORMULA:
{{{a=sqrt(2)}}}, {{{b=-3}}}, and {{{c=sqrt(2)}}}
{{{x=(-(-3) +- sqrt((-3)^2-4*sqrt(2)*sqrt(2)))/2sqrt(2)}}}-->{{{x=(3 +- sqrt(9-4*2))/2sqrt(2)}}}
{{{x=(3 +- sqrt(9-8))/2sqrt(2)}}}-->{{{x=(3 +- sqrt(1))/2sqrt(2)}}}-->{{{x=(3 +- 1)/2sqrt(2)}}}
The two solutions are {{{x=4sqrt(2)/2=2/sqrt(2)}}} and {{{x=2sqrt(2)/2=1/sqrt(2)}}}
Let's multiply numerator and denominator times {{{sqrt(2)}}}
because teachers do not like square roots in denominators
{{{x=1/sqrt(2)}}}-->{{{x=sqrt(2)/sqrt(2)sqrt(2)}}}-->{{{highlight(x=sqrt(2)/2)}}} and {{{x=2/sqrt(2)}}}-->{{{x=2sqrt(2)/sqrt(2)sqrt(2)}}}-->{{{x=2sqrt(2)/2}}}-->{{{highlight(x=sqrt(2))}}}
COMPLETING THE SQUARE:
I would like to make the equation a little simpler and easier to work with.
{{{sqrt(2)x^2-3x+sqrt(2)=0}}} --> {{{sqrt(2)*(sqrt(2)x^2-3x+sqrt(2))=sqrt(2)*0}}} --> {{{2x^2-3sqrt(2)x+2=0}}}
{{{2x^2-3sqrt(2)x+2=0}}}--> {{{x^2-(3sqrt(2)/2)x+1=0}}}(dividing both sides of the equation by 2) 
{{{x^2-(3sqrt(2)/2)x+1=0}}}-->{{{x^2-(3sqrt(2)/2)x=-1}}}
{{{x^2-(3sqrt(2)/2)x+(3sqrt(2)/4)^2=-1+(3sqrt(2)/4)^2}}}-->{{{(x-(3sqrt(2)/4))^2=-1+3^2(sqrt(2))^2/4^2}}}-->{{{(x-(3sqrt(2)/4))^2=-1+9*2/16}}}
{{{(x-(3sqrt(2)/4))^2=-1+18/16}}}-->{{{(x-(3sqrt(2)/4))^2=2/16}}}
The two solutions would come from {{{x-(3sqrt(2)/4)=sqrt(2/16)}}} and {{{x-(3sqrt(2)/4)=-sqrt(2/16)}}}
{{{x-3sqrt(2)/4=sqrt(2/16)}}}-->{{{x-(3sqrt(2)/4)=sqrt(2)/4)}}}-->{{{x=3sqrt(2)/4+sqrt(2)/4)}}}-->{{{x=4sqrt(2)/4)}}}-->{{{highlight(x=sqrt(2))}}}
{{{x-3sqrt(2)/4=-sqrt(2/16)}}}-->{{{x-(3sqrt(2)/4)=-sqrt(2)/4)}}}-->{{{x=3sqrt(2)/4-sqrt(2)/4)}}}-->{{{x=2sqrt(2)/4)}}}-->{{{highlight(x=sqrt(2)/2)}}}
FACTORING:
Starting from the simpler {{{x^2-(3sqrt(2)/2)x+1=0}}}
I would look for two numbers whose product is 1 and whose sum is {{{-(3sqrt(2)/2)}}}.
They must be both negative and there is a {{{sqrt(2)}}} in there somewhere.
Since the product is 1, one number is the reciprocal of the other, like {{{sqrt(2)}}} and {{{1/sqrt(2)}}}.
Trying {{{-sqrt(2)}}} and {{{-1/sqrt(2)}}}, I find that those numbers work
{{{-1/sqrt(2)=-1*sqrt(2)/sqrt(2)sqrt(2)=-sqrt(2)/2}}},
The sum is:
{{{-sqrt(2)+(-sqrt(2)/2)=-2sqrt(2)/2-sqrt(2)/2=-3sqrt(2)/2}}}
The product is
{{{-sqrt(2)*(-sqrt(2)/2)=sqrt(2)sqrt(2)/2)=2/2=1}}}
So the factoring gives me
{{{x^2-(3sqrt(2)/2)x+1=(x-sqrt(2))(x-sqrt(2)/2)}}}
meaning that thw equation can be written as
{{{(x-sqrt(2))(x-sqrt(2)/2)=0}}}  and the solutions are
{{{highlight(x=sqrt(2))}}} and {{{highlight(x=sqrt(2)/2)}}}