Question

If $x= \frac{ \sqrt{a+2b}- \sqrt{a-2b} }{\sqrt{a+2b}- \sqrt{a-2b} }$
then prove
$bx^2 -ax+b=0$
URGENT!!!!!!!!!! PLEASE

Answer 1

$x = \frac{ \sqrt{a+2b}+ \sqrt{a-2b} }{\sqrt{a+2b}- \sqrt{a-2b} } \\ \\ = \frac{ (\sqrt{a+2b}+ \sqrt{a-2b})(\sqrt{a+2b}+ \sqrt{a-2b}) }{(\sqrt{a+2b}- \sqrt{a-2b})(\sqrt{a+2b}+ \sqrt{a-2b}) } \\ \\ = \frac{a+2b + a-2b + 2 \sqrt{a+2b}\sqrt{a-2b} }{a+2b - (a-2b)} \\ \\ = \frac{2a + 2\sqrt{a^2-4b^2}}{4b}\\ \\x\ =\ \frac{a + \sqrt{a^2-4b^2}}{2b}$

This is in the format of a root of a quadratic equation as : 
$Roots\ of\ ax^2+bx+c = 0\ are\ x\ = \frac{-b +- \sqrt{b^2-4ac}}{2a}$

So use similarity:  a becomes -b;  c becomes a = -b   as,   4b² = 4 a c

So  -b x² + a x - b = 0

b x² - ax +b = 0

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Or, another way,

2 b x = a + √(a²-4b²)

(2bx-a)² = a² - 4b²
4b²x² +a² - 4abx = a² - 4b²
4b²x² - 4abx + 4b² = 0

divide by 4b

bx² - ax +b = 0

Answer 2

see the attachment for the soln.