SOLUTION: I have two questions both on quadratic equations. I am not understanding how to do them. The first one says to write it in standard quadratic form. The solutions are negative sig

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I have two questions both on quadratic equations. I am not understanding how to do them. The first one says to write it in standard quadratic form. The solutions are negative sig      Log On


   

Question 540037: I have two questions both on quadratic equations. I am not understanding how to do them. The first one says to write it in standard quadratic form. The solutions are
negative sign sqrt 5, and 4 sqrt 5.
The second one says to find the solution(s).
w^4-20w^2-2=0.
If you could help with either of these I would greatly appriciate it.
Thanks so much!
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
The first one says to write it in standard quadratic form. The solutions are
negative sign sqrt 5, and 4 sqrt 5.
:
x = -sqrt%285%29 is derived from the factor (x + sqrt%285%29)
and
x = 4sqrt%285%29 is derived from the factor (x - 4sqrt%285%29)
FOIL
(x - 4sqrt%285%29)(x + sqrt%285%29) = x^2 + sqrt%285%29x - 4sqrt%285%29x - 4(5)
:
x^2 - 3sqrt%285%29x - 20, is the quadratic equation for these solutions
:
:
The second one says to find the solution(s).
w^4 - 20w^2 - 2 = 0.
Use the quadratic formula to find w^2
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x = w^2, a=1, b=-20, c=-2
w%5E2+=+%28-%28-20%29+%2B-+sqrt%28-20%5E2-4%2A1%2A-2+%29%29%2F%282%2A1%29+
:
w%5E2+=+%2820+%2B-+sqrt%28400%2B8+%29%29%2F2+
w%5E2+=+%2820+%2B-+sqrt%28408+%29%29%2F2+
Two solutions
w%5E2+=+%2820+%2B+20.2%29%2F2+
w%5E2+=+40.2%2F2
w%5E2+=+20.1
w%5E2 = +/- sqrt%2820.1%29
w = +4.4833
and
w = -4.4833, these are only real solutions to this
the other solution will be square root of a negative number