Solve using the quadratic equation - 2r^4 - r^2 - 5 = 0? I WILL Choose a best answer!?

Question

^ means square root. So 2r^4 means 2r to the 4th power. Solve for R people! Or do whatever!

Thats how its written.... 2r^4

Not negative sorry that was just a dash, it's just

2r^4 - r^2 - 5 = 0

Answer 1

agree with larry,, anywway,, and is 2r a negative?

2r^4 - r^2 - 5 = 0 *remove 4*
r^4 - r^2 / 2 = 5 / 2 *transfer constant*
r^4 - r^2 / 2 + 1/16 = 5/2 + 1/16 *complete the square (b/2)^2*
(r^2 + 1/4)^2 = 41/16 (get the square root)
r^2 + 1/4 = plus-minus, square root of 41/4 *square root*
r^2 = -1 plus-minus, square root of 41 all over 4

eek didnt saw the ^2.. just do the rest ;)

Answer 2

2r^4 does not mean 2r is raised to the 4th power. It means r is raised to the 4th power and that is multiplied by 2. If you want 2r raised to the 4th power, then it must be written (2r)^4. So, which one is it in your problem? Until I know for sure I will not spend time helping out.

Answer 3

Is that 2r^4 OR -2r^4 ?
Without that, a specific answer becomes impossible.
The GENERAL answer is:

(2r^2 +A) * (r^2 +B)
with the possibility of ONE one the r^2 being negative.
A*B = -5
IF both r^2 are positive, then:
A+2B = -1
Otherwise:
-A+2B = -1 OR A-2B =-1

Answer 4

let x=r^2 and then re write your equation as
-2(x^2)- x- 5. Now you can use the quadratic formula.
and get x
x=r^2
r would be +/- (sqrt(the above answer)