Equation With Proportional Roots?
a + 3 a = 6 6 a
Find the number of solutions of a for the above equation.
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1 solution
Moderator note:
Great problem!
As always, when manipulating an equation, we have to careful and check that we didn't introduce extraneous solutions. In this case, we could better justify the substitution by saying "The domain of a is [ 0 , ∞ ) . Thus, we can use the bijective substitution a = x 6 , x ∈ 9 0 , ∞ ) . This explicitly explains how to deal with the solutions of the equation in x .
@Otto Bretscher Even I am back to roots! :P
Exactly Same Way.
I think this is wrong because (-3)^6 of course satisfies the equation and a polynom 3rd grades ALWAYS has three roots - in our case one real and two (conjugate) complex ones!!!
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Can you show your working how ( − 3 ) 6 satisfies the equation?
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Of course: ((-3)^6)^0.5+((-3)^6)^(1/3)=(-3)^3+(-3)^2=-27+9= -18=6*((-3)^6)^(1/6)=-18 q.e.d.
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@Andreas Wendler – 6 ( − 3 ) 6 = − 3 .
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@Nihar Mahajan – Surely this is: ((-3)^6)^(1/6)=(-3)^(6*1/6))=(-3)^1=-3 ;-)
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@Andreas Wendler – You are not allowed to take square root like that. x 2 = ∣ x ∣ and not x . You may read this wiki
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@Nihar Mahajan – The range of the solutions of your task is NOT restricted to real numbers! Have you ever heard about complex numbers?
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@Andreas Wendler – Do you know the proper definition of taking a square root?
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@Nihar Mahajan – sqrt(1)=1 and sqrt(-1)=i. ;-)))
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@Andreas Wendler – Fine , since you are using complex numbers , have a look at this:
6 ( − 3 ) 6 = 6 3 6 i 1 2 = 6 3 6 × 1 = 3 × 1 = 3
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@Nihar Mahajan – i 1 2 = + 1 < > − 1 = > Y o u f o o l !
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@Andreas Wendler – Unfortunately you are absolutely out of idea. That's why I advice the following link for you:
https://www.youtube.com/watch?v=JdbZYrxL5OU
@Nihar Mahajan – − 3 = 3 e i π = > ( − 3 ) 6 = 3 6 e 6 i π = > ( − 3 ) 6 ) 6 1 = 3 e i π = − 3
Other questions?
For clarity, you might want to write "Find the number of solutions a ≥ 0 for the above equation" since your solution does not apply to negative a .
First substitute a = x 6 to get cleaner form of the equation:
x 3 + x 2 = 6 x ⇒ x 3 + x 2 − 6 x = 0 ⇒ x ( x 2 + x − 6 ) = 0 ⇒ x ( x − 2 ) ( x + 3 ) = 0
Now this gives us x = { 0 , 2 , − 3 } ⇒ a = { 0 , 6 4 , ( − 3 ) 6 } . But why is the answer 2 ?
This is because when a = ( − 3 ) 6 does not satisfy the given equation and is kinda extraneous solution. Hence a = { 0 , 6 4 } and only 2 solutions for a in the equation.