Can someone check my work for this quadratic equation? Thank you.
b^4 + 2b^2 - 24 = 0
b^4 + 2b^2 = 24
(b^2)^2 + 2b^2 = 24
b^2 + 2b = 24
....
b = -2 +(-) ( Radical 4 - 4 (1)(-24) ) / 2(1)
b = -2 +(-) (Radical 100) /2
b = - 2 +(-) 10 / 2
b = -1 +(-) 5
b1 = 4 b2= -6
3 answers
Someone said I got it right, but then the response got deleted. Does that mean my answer is incorrect?
a 4th order equation usually has 4 solutions
b^4 + 2b^2 - 24 = 0 ... (b^2 + 6)(b^2 - 4) = 0
b^2 + 6 = 0 ... b = ± i √6
b^2 - 4 = 0 ... b = ± 2
b^4 + 2b^2 - 24 = 0 ... (b^2 + 6)(b^2 - 4) = 0
b^2 + 6 = 0 ... b = ± i √6
b^2 - 4 = 0 ... b = ± 2
suppose you let x = b^2
then your equation becomes
x^2 + 2x - 24 = 0
This factors nicely to
(x + 6)(x - 4) = 0
x = -6 or x = 4
then b^2 = -6 or b^2 = 4
let's handle the b^2 = 4
clearly b = ± 2
what about the b^2 = -6
there is no solution in the set of real numbers, since squaring any number would make it positive.
If you have studied complex numbers, then
b = ± i√6 , were i is defined as √-1
as to your solution, you could have done:
(b^2 + 6)(b^2 - 4) = 0
to get b^2 = -6 or b^2 = 4 , same as my result above.
your step from
(b^2)^2 + 2b^2 = 24
to
b^2 + 2b = 24
makes no sense.
then your equation becomes
x^2 + 2x - 24 = 0
This factors nicely to
(x + 6)(x - 4) = 0
x = -6 or x = 4
then b^2 = -6 or b^2 = 4
let's handle the b^2 = 4
clearly b = ± 2
what about the b^2 = -6
there is no solution in the set of real numbers, since squaring any number would make it positive.
If you have studied complex numbers, then
b = ± i√6 , were i is defined as √-1
as to your solution, you could have done:
(b^2 + 6)(b^2 - 4) = 0
to get b^2 = -6 or b^2 = 4 , same as my result above.
your step from
(b^2)^2 + 2b^2 = 24
to
b^2 + 2b = 24
makes no sense.
1 answer