Math Is Fun Forum
Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °
You are not logged in.
- Topics: Active | Unanswered
#1 2006-12-26 03:58:30
- Neha
- Member
- Registered: 2006-10-11
- Posts: 173
Help
Write the quadratic equation with a lead coefficient of 1 whose roots are 2 + sqrt5 and 2 - sqrt5 .
Live Love Life
Offline
#2 2006-12-26 06:17:13
- mathsyperson
- Moderator
- Registered: 2005-06-22
- Posts: 4,900
Re: Help
If the lead coefficient is 1, then by the quadratic equation, the solution will be x = [-b ±√(b² - 4c)]/2.
We are told that the roots are 2+sqrt5 and 2-sqrt5, and this can be rewritten as [4±√20]/2.
By equating these, we get that -b = 4 and b² - 4c = 20.
From the first equation, b = -4. Substituting this into the second equation gets 16 - 4c = 20 and hence c = -1.
Therefore, the quadratic equation that gives these roots is x² - 4x - 1 = 0.
Why did the vector cross the road?
It wanted to be normal.
Offline