f(sqrt 3-sqrt 2)=x^2-3x+4
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y =(3^.5-2^.5)^2 - 3(3^.5-2^.5) + 4
= 3 - 2(6^.5) + 2 - 2(3^.5) - 3(2^.5) + 4
= 9 - 2(6^.5) - 3^(1.5) - 3(2^.5)
To solve the equation f(sqrt(3) - sqrt(2)) = x^2 - 3x + 4, we need to first understand what the function f() represents. We'll assume that f() is a function that takes a numerical value as input and returns another numerical value as output.
Let's start by evaluating the expression inside f(). We have f(sqrt(3) - sqrt(2)).
To simplify this expression, we use the difference of squares formula: a^2 - b^2 = (a + b)(a - b).
In our case, a = sqrt(3) and b = sqrt(2). Plugging these values into the formula, we get:
(sqrt(3))^2 - (sqrt(2))^2 = (sqrt(3) + sqrt(2))(sqrt(3) - sqrt(2)).
Simplifying further, we have:
3 - 2 = (sqrt(3) + sqrt(2))(sqrt(3) - sqrt(2)).
1 = (sqrt(3) + sqrt(2))(sqrt(3) - sqrt(2)).
Now we can substitute this value back into the original equation:
f(sqrt(3) - sqrt(2)) = x^2 - 3x + 4.
f(1) = x^2 - 3x + 4.
The equation becomes:
x^2 - 3x + 4 = 0.
Now we have a quadratic equation. To solve it, we can use methods like factoring, completing the square, or using the quadratic formula.
Let's assume we solve it using the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / (2a),
where in our case, a = 1, b = -3, and c = 4.
Plugging in these values, we have:
x = (-(-3) ± sqrt((-3)^2 - 4(1)(4))) / (2(1)),
x = (3 ± sqrt(9 - 16)) / 2,
x = (3 ± sqrt(-7)) / 2.
Since we have a square root of a negative number, the equation does not have any real solutions. Therefore, there is no real value of x that satisfies the equation x^2 - 3x + 4 = 0 after evaluating f(sqrt(3) - sqrt(2)).