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lgraha13 lgraha13
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11 years ago
I know how to solve quadratics, but this one has stumped me because
I don't know how to reduce the exponents..
How do I work this problem out?
I am frustrated because I can't find help online anywhere :/

w^4+7w^2-18=0.
Thanks for the help everyone!
As I read the explanations everyone provided, I slapped my forehead. Duh!
I remember working problems out like this last week.
I have finals tomorrow and so, I am just trying to go over everything I do not understand one last time.
If you could answer some other problems I posted up, I would really appreciate it.
<3
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wrote...
#1
11 years ago
Let t = w^2 then it just becomes a basic quadratic

t^2 + 7t - 18 = 0
(t - 2)(t + 9) = 0

Then just sub in w^2 = t, continue on and thats it Slight Smile
wrote...
#2
11 years ago
With substitution it becomes a quadratic.  Let u = w² and rewrite as:
u² + 7u - 18 = 0
(u+9)(u-2) = 0
u  = -9 and 2 = w²

w = ±3i and ±?2
wrote...
#3
11 years ago
w^4+7w^2-18=0
(w^2)^2+7w^2-18=0
let u=x^2
u^2+7u-18=0
then use quadratic equation. just remember that when you get an answer, it is the value for w^2, you have to do one extra step square root it.
wrote...
#4
11 years ago
define y = w^2

then your problem reduces to

y^2+7y-18=0

y will be  (-7+(49+73)^0.5)/2=2.02

so w= -1.41 , 1.41

only two real roots
wrote...
#5
11 years ago
Same as any quadratic equation.
(w^2 - 2)(w^2 + 9) = 0
You cannot reduce this any further using whole numbers
So w^2 - 2 = 0
w^2 = 2
(w^2 - 2) could be reduced to (w + sqrt 2)(w - sqrt 2)
w = +/-sqrt 2
w^2 + 9 = 0
w^2 = -9
w = 3 sqrt (-1)
wrote...
#6
11 years ago
Quartic equations have quadratic factors.
w? + 7w² - 18 = 0 can be factored as (w² - 2)(w² + 9) = 0.
This means that w² - 2 = 0 or w² + 9 = 0.
w² - 2 = 0 means that w² = 2, so w = ±?2.
w² + 9 = 0 means that w² = -9, so w = ±?(-9) = ±3i.
The solution set is {?2, -?2, 3i, -3i}.
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