Substitute x in an exponent
x 5 + x 4 + x 2 + x = 3 3 6
Find the positive value of x that satisfies the equation above.
The answer is 3.
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8 solutions
Moderator note:
Almost complete. You need to explain why then quartic factor does not have a positive root.
Hint : Descartes' Rule of Sign.
There's no sign change in the quartic factor that's why it doesn't have a positive root.
Note that x = 3 is a solution. Let f ( x ) = x 5 + x 4 + x 2 + x − 3 3 6 . We get f ′ ( x ) = 5 x 4 + 4 x 3 + 2 x + 1 . Note that f ′ ( x ) > 0 for all x ≥ 0 . Hence, the function is strictly increasing for x ≥ 0 . We get x = 3 as our unique positive solution.
Moderator note:
Yes. This is the calculus approach.
x 5 + x 4 + x 2 + x = 3 3 6 ( x 5 + x 2 ) + ( x 4 + x ) = 3 3 6 x 2 ( x 3 + 1 ) + x ( x 3 + 1 ) = 3 3 6 x ( x 3 + 1 ) ( x + 1 ) = 3 3 6 x ( x + 1 ) 2 ( x 2 − x + 1 ) = 3 3 6 … ( ∗ ) 3 3 6 can be factor into 3 3 6 = 3 . 2 4 . 7 = 3 . 4 2 . 7 = e l s e so, for ( ∗ ) we can chose x ( x + 1 ) 2 ( x 2 − x + 1 ) = 3 . 4 2 . 7 because x = 3 ( x + 1 ) 2 = ( 3 + 1 ) 2 = 4 2 x 2 − x + 1 = 3 2 − 3 + 1 = 7 x = 3
Moderator note:
Unless we know beforehand that there is only one positive solution of x , your solution is not complete. This is because you have only shown that x = 3 is a positive solution. How do you know that there can't be any other positive solution?
How do you know that x is an integer?
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From x 5 + x 4 + x 2 + x − 3 3 6 = 0 , we can chose solution x using combination f a c t o r 1 f a c t o r 3 3 6 . it means solution x is an integer
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I am sorry I did not understand your comment. How could you be sure that x cannot be a non-integer?
L e t f ( x ) = x 5 + x 4 + x 2 + x − 3 3 6 = 0 . 3 3 6 = 2 4 ∗ 3 ∗ 7 . ∴ x = 2 n / 3 / 7 . n = 1 , 2 , 3 , o r 4 . f ( 2 ) = 0 . f ( 3 ) = 0 . ∴ x = 3 By synthetic division, we get , f ( x ) = ( x − 3 ) ∗ ( x 4 + 4 x 3 + 1 2 x 2 + 3 7 x + 1 1 2 ) = 0 , In the remaining quartic , there is no change of sign. So no othere + tive roots as per Descartes’ Rule of Sign
Is it correct to say that x one of the factors of 336? since x is a factor of the given expression?
Moderator note:
The question did not explicitly mention that x must be an integer. How do you know that there can't be a positive non-integer solution?
Thank you for your pointing out. I have made additions to complete the solution.
x 5 + x 4 + x 2 + x x 4 ( x + 1 ) + x ( x + 1 ) x ( x 3 + 1 ) ( x + 1 ) x ( x + 1 ) 2 ( x 2 − x + 1 ) 3 ( 3 + 1 ) 2 ( 3 2 − 3 + 1 ) ⇒ x = 3 3 6 = 3 × 2 4 × 7 = 3 × 2 4 × 7 = 3 × 2 4 × 7 = ( 3 ) ( 4 2 ) ( 7 ) [ By observation ] = 3
Factorizing x 5 + x 4 + x 2 + x − 3 3 6 = 0 , we have:
( x − 3 ) ( x 4 + 4 x 3 + 1 2 x 2 + 3 7 x + 1 1 2 ) = 0
The equation has only one real root x = 3 .
Moderator note:
Hm, that observation is almost magical. How else can we motivate it?
by looking at it you can tell it is a multiple of 3, so it has the possibility of being an answer relating to the low multiples of 3, hence 3, 6, 9, 12. however by using trail and error youd see that 6 is way too high and so must be 3
you can solve this by trial and error...
lets take 2 ,answer is 44 then go for 3 its 336, very easy.
333 + 3 + 336 ;-) that's how I did it haha
x 5 + x 4 + x 2 + x = 3 3 6 ⇔ x 5 + 4 x 4 − 3 x 4 + 1 2 x 3 − 1 2 x 3 + 3 7 x 2 − 3 6 x 2 + 1 1 2 x − 1 1 1 x − 3 3 6 = 0 ⇔ ( x − 3 ) ( x 4 + 4 x 3 + 1 2 x 2 + 3 7 x + 1 1 2 ) = 0 ⇒ x − 3 = 0 ⇒ x = 3