Quandaries
and Queries
my name is Bob.
while looking at x=(y2-b2)/2b on a table I was preparing for a graph, I saw that there is a point where x and y get closer to each other's value. By trial and error I tried to find the point where x=y. I saw a number which looked familiar and I realized it was the square root of 2. and so, if y=b(1+square root of 2) then x=b(1+square root of 2) also, or as x=y.
My question is: How can this result x=y when y=b(1+square root of 2) be algebraically derived from the equation x=(y2-b2)/2b without pre knowing it is so?
Thanks
Bob
Hi Bob,
You are looking for the value of x when x = y. When x = y your equation becomes
x=(x2-b2)/2b
or, after simplifying
x2 - 2bx - b2 = 0
This quadratic in x doesn't factor but you can solve it using the general quadratic. When you do so you will find, not only
x = b( 1 + √2)
but also
x = b( 1 - √2)
Penny
