SOLUTION: 4x^2-4sq.root3x+3=0 sq.root2x^2-3x+sq.root2=0

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Question 633226: 4x^2-4sq.root3x+3=0
sq.root2x^2-3x+sq.root2=0

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
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%5E2-4sqrt%283%29x%2B3=0
SMART FACTORING:
4x%5E2=%282x%29%5E2 is a square.
I would say that 3=%28-sqrt%283%29%29%5E2 is another square.
Is -4sqrt%283%29x twice the product of 2x and -sqrt%283%29?
Yes, 2%2A%282x%29%2A%28-sqrt%283%29%29=-4sqrt%283%29x .
Then, 4x%5E2-4sqrt%283%29x%2B3 is the square of a binomial.
%282x-sqrt%283%29%29%5E2=4x%5E2-4sqrt%283%29x%2B3
The equation can be written as
%282x-sqrt%283%29%29%5E2=0 --> 2x-sqrt%283%29=0 --> highlight%28x=sqrt%283%29%2F2%29
COMPLETING THE SQUARE:
4x%5E2-4sqrt%283%29x%2B3=0 --> x%5E2-sqrt%283%29x%2B3%2F4=0 --> x%5E2-sqrt%283%29x=-3%2F4
Adding 3%2F4 (the square of -sqrt(3)/2) to both sides we form the square of x-sqrt%283%29%2F2 on the left side:
x%5E2-sqrt%283%29x=-3%2F4 --> x%5E2-sqrt%283%29x%2B3%2F4=-3%2F4%2B3%2F4 --> %28x-sqrt%283%29%2F2%29%5E2=0 --> x-sqrt%283%29%2F2=0 --> highlight%28x=sqrt%283%29%2F2%29
USING THE QUADRATIC FORMULA:
For the generic quadratic equation ax%5E2%2Bbx%2Bc=0 the solution(s) is/are
x=%28-b+%2B-+sqrt%28b%5E2-4ac%29%29%2F2a (if what's under the square root is not negative)
In the case of 4x%5E2-4sqrt%283%29x%2B3=0 , a=4, b=-4sqrt%283%29, and c=3
Substituting:
x=%28-%28-4sqrt%283%29%29+%2B-+sqrt%28%28-4sqrt%283%29%29%5E2-4%2A4%2A3%29%29%2F%282%2A4%29 --> x=%284sqrt%283%29+%2B-+sqrt%28%28-4%29%5E2%28sqrt%283%29%29%5E2-48%29%29%2F8

x=%284sqrt%283%29+%2B-+sqrt%2816%2A3-48%29%29%2F8 --> x=%284sqrt%283%29+%2B-+sqrt%2848-48%29%29%2F8

x=%284sqrt%283%29+%2B-+sqrt%280%29%29%2F8 --> x=%284sqrt%283%29+%2B-+sqrt%280%29%29%2F8

x=%284sqrt%283%29+%2B-+0%29%2F8 --> x=4sqrt%283%29%2F8 --> highlight%28x=sqrt%283%29%2F2%29

sqrt%282%29x%5E2-3x%2Bsqrt%282%29=0
USING THE QUADRATIC FORMULA:
a=sqrt%282%29, b=-3, and c=sqrt%282%29
x=%28-%28-3%29+%2B-+sqrt%28%28-3%29%5E2-4%2Asqrt%282%29%2Asqrt%282%29%29%29%2F2sqrt%282%29-->x=%283+%2B-+sqrt%289-4%2A2%29%29%2F2sqrt%282%29
x=%283+%2B-+sqrt%289-8%29%29%2F2sqrt%282%29-->x=%283+%2B-+sqrt%281%29%29%2F2sqrt%282%29-->x=%283+%2B-+1%29%2F2sqrt%282%29
The two solutions are x=4sqrt%282%29%2F2=2%2Fsqrt%282%29 and x=2sqrt%282%29%2F2=1%2Fsqrt%282%29
Let's multiply numerator and denominator times sqrt%282%29
because teachers do not like square roots in denominators
x=1%2Fsqrt%282%29-->x=sqrt%282%29%2Fsqrt%282%29sqrt%282%29-->highlight%28x=sqrt%282%29%2F2%29 and x=2%2Fsqrt%282%29-->x=2sqrt%282%29%2Fsqrt%282%29sqrt%282%29-->x=2sqrt%282%29%2F2-->highlight%28x=sqrt%282%29%29
COMPLETING THE SQUARE:
I would like to make the equation a little simpler and easier to work with.
sqrt%282%29x%5E2-3x%2Bsqrt%282%29=0 --> sqrt%282%29%2A%28sqrt%282%29x%5E2-3x%2Bsqrt%282%29%29=sqrt%282%29%2A0 --> 2x%5E2-3sqrt%282%29x%2B2=0
2x%5E2-3sqrt%282%29x%2B2=0--> x%5E2-%283sqrt%282%29%2F2%29x%2B1=0(dividing both sides of the equation by 2)
x%5E2-%283sqrt%282%29%2F2%29x%2B1=0-->x%5E2-%283sqrt%282%29%2F2%29x=-1
x%5E2-%283sqrt%282%29%2F2%29x%2B%283sqrt%282%29%2F4%29%5E2=-1%2B%283sqrt%282%29%2F4%29%5E2-->%28x-%283sqrt%282%29%2F4%29%29%5E2=-1%2B3%5E2%28sqrt%282%29%29%5E2%2F4%5E2-->%28x-%283sqrt%282%29%2F4%29%29%5E2=-1%2B9%2A2%2F16
%28x-%283sqrt%282%29%2F4%29%29%5E2=-1%2B18%2F16-->%28x-%283sqrt%282%29%2F4%29%29%5E2=2%2F16
The two solutions would come from x-%283sqrt%282%29%2F4%29=sqrt%282%2F16%29 and x-%283sqrt%282%29%2F4%29=-sqrt%282%2F16%29
x-3sqrt%282%29%2F4=sqrt%282%2F16%29-->x-%283sqrt%282%29%2F4%29=sqrt%282%29%2F4%29-->x=3sqrt%282%29%2F4%2Bsqrt%282%29%2F4%29-->x=4sqrt%282%29%2F4%29-->highlight%28x=sqrt%282%29%29
x-3sqrt%282%29%2F4=-sqrt%282%2F16%29-->x-%283sqrt%282%29%2F4%29=-sqrt%282%29%2F4%29-->x=3sqrt%282%29%2F4-sqrt%282%29%2F4%29-->x=2sqrt%282%29%2F4%29-->highlight%28x=sqrt%282%29%2F2%29
FACTORING:
Starting from the simpler x%5E2-%283sqrt%282%29%2F2%29x%2B1=0
I would look for two numbers whose product is 1 and whose sum is -%283sqrt%282%29%2F2%29.
They must be both negative and there is a sqrt%282%29 in there somewhere.
Since the product is 1, one number is the reciprocal of the other, like sqrt%282%29 and 1%2Fsqrt%282%29.
Trying -sqrt%282%29 and -1%2Fsqrt%282%29, I find that those numbers work
-1%2Fsqrt%282%29=-1%2Asqrt%282%29%2Fsqrt%282%29sqrt%282%29=-sqrt%282%29%2F2,
The sum is:
-sqrt%282%29%2B%28-sqrt%282%29%2F2%29=-2sqrt%282%29%2F2-sqrt%282%29%2F2=-3sqrt%282%29%2F2
The product is
-sqrt%282%29%2A%28-sqrt%282%29%2F2%29=sqrt%282%29sqrt%282%29%2F2%29=2%2F2=1
So the factoring gives me
x%5E2-%283sqrt%282%29%2F2%29x%2B1=%28x-sqrt%282%29%29%28x-sqrt%282%29%2F2%29
meaning that thw equation can be written as
%28x-sqrt%282%29%29%28x-sqrt%282%29%2F2%29=0 and the solutions are
highlight%28x=sqrt%282%29%29 and highlight%28x=sqrt%282%29%2F2%29