Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#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

Board footer

Powered by FluxBB