If
x= \frac{ \sqrt{a+2b} + \sqrt{a-2b} }{ \sqrt{a+2b} - \sqrt{a-2b} }
Show that: bx^2-ax+b=0
To show that bx^2 - ax + b = 0, we need to start with the expression for x and simplify it.
Given: x = (sqrt(a + 2b) + sqrt(a - 2b)) / (sqrt(a + 2b) - sqrt(a - 2b))
To eliminate the square roots in the denominator, we can multiply the numerator and denominator by the conjugate of the denominator:
x = (sqrt(a + 2b) + sqrt(a - 2b)) / (sqrt(a + 2b) - sqrt(a - 2b)) * (sqrt(a + 2b) + sqrt(a - 2b)) / (sqrt(a + 2b) + sqrt(a - 2b))
This simplifies to:
x = (a + 2b) + sqrt((a + 2b)(a - 2b)) / (a + 2b) - (a - 2b)
Now, let's simplify the expression further:
x = (a + 2b + sqrt(a^2 - 4b^2)) / (a + 2b - a + 2b)
x = (2(a + 2b) + sqrt(a^2 - 4b^2)) / (4b)
Next, let's rewrite x in terms of the original expression by multiplying both the numerator and denominator by 2:
x = (4(a + 2b) + 2sqrt(a^2 - 4b^2)) / (8b)
Now, simplify the expression further:
x = (2(a + 2b) + sqrt(a^2 - 4b^2)) / (4b)
x = a/2b + 1 + (sqrt(a^2 - 4b^2)) / (4b)
Comparing this expression with bx^2 - ax + b, we can identify the coefficients, which are:
a = 1
b = 1/4
Therefore, bx^2 - ax + b = (1/4)x^2 - x + 1/4 = 0, as required.