Roots of Quadratic Equation

**magentarita** · January 4th 2009, 12:51 PM

Which term describes the roots of the equation
2x^2 + 3x - 1 = 0?

(1) rational (3) equal
(2) irrational (4) imaginary

I guessed choice (2) and was right but I still do not know what makes the roots irrational in this case.

**nzmathman** · January 4th 2009, 01:05 PM

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.

**magentarita** · January 5th 2009, 09:53 PM

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.

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nice....

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