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

Question 633226: 4x^2-4sq.root3x+3=0
sq.root2x^2-3x+sq.root2=0
Answer by KMST(5328) (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.

SMART FACTORING:
is a square.
I would say that is another square.
Is twice the product of and ?
Yes, .
Then, is the square of a binomial.

The equation can be written as
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COMPLETING THE SQUARE:
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Adding (the square of -sqrt(3)/2) to both sides we form the square of on the left side:
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USING THE QUADRATIC FORMULA:
For the generic quadratic equation the solution(s) is/are
(if what's under the square root is not negative)
In the case of , , , and
Substituting:
-->

-->

-->

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USING THE QUADRATIC FORMULA:
, , and
-->
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The two solutions are and
Let's multiply numerator and denominator times
because teachers do not like square roots in denominators
-->--> and -->-->-->
COMPLETING THE SQUARE:
I would like to make the equation a little simpler and easier to work with.
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--> (dividing both sides of the equation by 2)
-->
-->-->
-->
The two solutions would come from and
-->-->-->-->
-->-->-->-->
FACTORING:
Starting from the simpler
I would look for two numbers whose product is 1 and whose sum is .
They must be both negative and there is a in there somewhere.
Since the product is 1, one number is the reciprocal of the other, like and .
Trying and , I find that those numbers work
,
The sum is:

The product is

So the factoring gives me

meaning that thw equation can be written as
and the solutions are
and

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