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There should be 3 solutions, I got 2... x = a - square root of x and x = a^2 - 2ax + x^2.

2007-10-22 10:39:11 · 1 answers · asked by amy 1 in Science & Mathematics Mathematics

1 answers

Why should there be three solutions? It's a quadratic in sqrt(x).

If you put sqrt(x) = y then the equation becomes

y^2 + y = a

or y^2 + y - a = 0

Then by the quadratic formula,

y = [-1 +/- sqrt(1 + 4a)]/2

x = y^2

So x = {[-1 +/- sqrt(1 + 4a)]/2}^2

= [-1 +/- sqrt(1 + 4a)]^2/4

= [1 +/- 2 sqrt(1 + 4a) + (1 + 4a)]/4

= [1 + 2a +/- sqrt(1 + 4a)]/2

2007-10-23 02:17:45 · answer #1 · answered by chauncy 7 · 0 0

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