Solve b^4+b^2-20=0?

Answer 1

First, replace b^2 with x and rewrite it as a quadratic:

x^2 + x - 20 = 0

This gives you:

(x +5)(x - 4) = 0.

Now, put the b^2 back in it.

(b^2 + 5)(b^2 - 4) = 0

This factors just a little further:

(b^2 + 5)(b + 2)(b - 2) = 0.

This has two real solutions:
b = 2, b = -2.

If you were studying complex arithematic, you would have 2 imaginary roots as well.

Answer 2

b=2

Answer 3

b= 2 or b= -2

Answer 4

+2 or -2

Answer 5

b^4 + b^2 – 20 = ( b^2 - 4 ) ( b^2 + 5 ) = 0
=> b^2 = 4 or b2 = -5.
The last equality is not possible for real numbers so,
b^2 = 4 or b = +2 and b = - 2

Answer 6

Treat this as c^2 + c - 20 where c = b^2.

Factor this as (c + 5)(c - 4).

So c = -5 and c = 4

So b^2 = -5 and b^2 = 4.

B^2 = 4 can have the roots 2 and -2.

B^2 = -5 has the root \/5 * i (sqrt(5)*i) and -\/5 * i

Answer 7

(b^2-4)(b^2+5)=0

if either part =0, then the whole thing equals zero, so if b is 2 or -2, the equation is zero.

Answer 8

b^4 + b^2 - 20
(b^2 + 5)(b^2 - 4)
(b^2 + 5)(b - 2)(b + 2)

ANS : isqrt(5), -isqrt(5), 2, or -2

Answer 9

b^4+b^2-20=0
b^2(b^2+1)=20
b^2=4
b= + or - 2

Answer 10

b^4 + b^2 - 20 = 0
b^4 + b^2 = 20

hmmm I don't know what a caret means.