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what is the value of a^4-a^3+a^2+2 if a^2+2=2a
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gyanshreshta,
this is polynomial (actually, a monomial since only one variable is there in it)of degree 4, that is the highest power is 4.
You have not stated a^4-a^3+a^2+2=0.
Since = 0 is not given, it is only a ploynomial and not an equation.
It has already been given a^2+2=2a.
Substitute it in the original polynomial.
a^4-a^3+a^2+2=a^4-a^3+2a.
To factorise this, take the a outside the bracket. You get
a (a^3-a^2+2).
PS:- If you want to solve a cubic equation (degree 3) or quartic equation (degree 4) I think there's no prescribed method. I haven't come across any. I would do it by 'Trial and success' method.
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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You can use that identity many times to reduce the equation's order.
For example, multiplying both sides by a² gives that a^4 + 2a^2 = 2a^3.
Therefore, a^4-a^3+a^2+2 = a^3-a^2+2.
Similarly, a^3 + 2a = 2a^2, and that means that a^3-a^2+2 = a^2-2a+2.
Then, because a^2+2 = 2a, that gives a^2+2-2a = 0.
Why did the vector cross the road?
It wanted to be normal.
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I think you guys are making this too hard. All he has to do is solve for
using the equation given to him:Then once you find
you just substitute it back into the first polynomial to get it's value.Wrap it in bacon
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Heh, fair point.
Although looking at it, that would return a complex value of a, so it wouldn't be quite as trivial as you suggest.
Why did the vector cross the road?
It wanted to be normal.
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I didn't even think about that, but you're right. I hope he doesn't mind finding 4th powers of complex numbers.
Wrap it in bacon
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It's not actually too bad. Solving the quadratic gives a = 1±i.
Therefore, a² = ±2i and so a^4 = -4.
a³ = ±2i(1±i) = ±2i - 2.
Plugging all of that in gives that a^4-a^3+a^2+2 = -4 -(±2i - 2) ±2i + 2
= (-4+2+2) + (±2 - ±2)i = 0.
Two methods, same answer. More or less equally strenuous as well.
Why did the vector cross the road?
It wanted to be normal.
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thanks mathsyperson.
i myself finally did it in another way.
ie
=^2(a^2-2a+2)+a(a^2-2a+2)+1(a^2-2a+2)
=a^2+a+1)(a^2-2a+2)
=a^2+a+1)*0
=0
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