Thread: Roots of Quadratic Equation
View Single Post
  #2  
Old January 4th, 2009, 02:05 PM
nzmathman nzmathman is offline
Member
  
Join Date: Dec 2008
Location: Auckland, New Zealand
Posts: 182
Country:
Thanks: 10
Thanked 89 Times in 85 Posts
nzmathman will become famous soon enough
Default
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.
Reply With Quote
The Following 2 Users Say Thank You to nzmathman For This Useful Post:
Donate to MHF