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magentarita · #3 January 5th, 2009, 11:53 PM

Quote:

Originally Posted by nzmathman

The roots are found by the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In your equation, $a = 2$ , $b = 3$ and $c = -1$ .

So $x = \frac{-3 \pm \sqrt{9 + 8}}{4} = \frac{-3 \pm \sqrt{17}}{4}$

This shows there are 2 real roots. Now 17 is a prime number - it can only be divided exactly by 1 and itself. The square root of any prime number is irrational - it cannot be expressed as a fraction. So the roots of the equation are therefore irrational.

Very nicely explained.