Sum of the reals
x ( x + 1 ) ( x + 2 ) ( x + 3 ) = 1 2 0
Find the sum of all the real roots of the equation above.
The answer is -3.
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11 solutions
You could have taken ( x + 3 ) = a and proceeded in a shorter way though.
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Why is the (x+3) before (x+1) as the first equation? Should give you wrong result
Nice skills :D
I'm sorry, I don't understand how you got from the second line to the third.
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I not able to understand second and third line
Nicely done ! I too did the same way but you use to make everthting a perfect square which I skipped by simply checking the discriminant. 😀😀
are negative numbers real numbers?
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Yes they are. But the roots of negative numbers (for instance − 3 9 ) and simplified expressions involving them (for instance − 3 + − 3 9 ) aren't.
Awesome Answer BRO..
Fabulous solution
x(x+1)(x+2)(x+3)=120=(2)(3)(4)(5). Then it's more clear that 2 is one root. Similary, x(x+1)(x+2)(x+3)=(-2)(-3)(-4)(-5). As -5<-4<-3<-2 then -5 is another root. Reducing the equation by Briot-Ruffini Algorithm we obtain (x²+3x+12)(x-2)(x+5)=0. As x²+3x+12 only have imaginary roots, the sum of all real roots is -5+2=-3.
I learned from your post about how the Briot-Ruffini Algorithm is analogous to synthetic division. Nice! :)
This is the easiest solution. Thank you!
I made same
Only 2 solutions possible for this equation,
where x=2 /-5
Hence their sum is, -5+2 = -3
More two are possible, though they will be imaginary. But you can ignore it. So only 2 real solutions are possible. :)
Easier than you thought:
Factorize 120:
1 2 0 = 2 ∗ 3 ∗ 4 ∗ 5
1 2 0 = − 2 ∗ − 3 ∗ − 4 ∗ − 5 .
So x can be either 2 or − 5
Nice observation of factorial 5.
x ( x + 1 ) ( x + 2 ) ( x + 3 ) = 1 2 0 ⇒ x 4 + 6 x 3 + 1 1 x 2 + 6 x − 1 2 0 = 0
Using Lodovico Ferrari’s Method, we can find the roots of this polynomial.
⇒ x = − 5 ∨ 2 ∨ 2 − 3 ± i 3 9
Since two of these roots are real, the answer is: − 5 + 2 = − 3
What is lodovico ferrari's method ?
Let the given equation be p(x) = 120. Then the quartic polynomial p(x) is symmetric in the line x = -1.5, and has roots at x = -3, -2, -1, and 0.
Furthermore, p(x) has the classic "w" shape, featuring
(1) a local maximum at x = -1.5, with p(-1.5) = 9/16;
(2) two global minima, one in (-3, -2) and the other in (-1, 0).
Thus for any k > 9/16, the equation p(x) = k has exactly two real solutions. Since these solutions are symmetric in the line x = -1.5, their sum is -3.
QED
x ( x + 1 ) ( x + 2 ) ( x + 3 ) = 1 2 0
x 4 + 6 x 3 + 1 1 x 2 + 6 x = 1 2 0
x 4 + 6 x 3 + 1 1 x 2 + 6 x − 1 2 0 = 0
( x − 2 ) ( x + 5 ) ( x 2 + 3 x + 1 2 ) = 0
x − 2 = 0
x = 2
x + 5 = 0
x = − 5
− 5 + 2 = − 3
We can ignore the other solutions deriving from x 2 + 3 x + 1 2 = 0 as they include imaginary numbers.
5! = 120
x+3 = 5, x = 2
since there are 4 terms, it wouldn't matter if all of them were negative. by choosing -5 for the first term, we can obtain the same result.
x ( x + 3 ) ( x + 1 ) ( x + 2 ) = 1 2 0
( x 2 + 3 x ) ( x 2 + 3 x + 2 ) = 1 2 0
( x 2 + 3 x ) 2 + 2 ( x 2 + 3 x ) + 1 = 1 2 1
( x 2 + 3 x + 1 ) 2 = ( ± 1 1 ) 2
x 2 + 3 x = − 1 2 o r x 2 + 3 x = 1 0
4 x 2 + 1 2 x = − 4 8 o r 4 x 2 + 1 2 x = 4 0
( 2 x + 3 ) 2 = − 3 9 o r ( 2 x + 3 ) 2 = 4 9
First equation not to be considered (always have imaginary roots)
2 x + 3 = ± 7
x = − 5 , 2