give me an A
The set of values of a such that the equation
x 3 ( x + 1 ) = 2 ( x + a ) ( x + 2 a )
has 4 (not necessarily distinct) real solutions, is [ m , n ] .
Find m + n .
The answer is 0.375.
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1 solution
How do we think of such factorization? Is it just practice or there is some trick in it.
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What I hoped was that the quartic could be factored as
( x 2 + m x + 2 a ) ( x 2 + n x − 2 a ) for some integers m , n .
I then expanded this to get
x 4 + ( m + n ) x 3 + m n x 2 + 2 ( m − n ) x − 4 a 2 ,
and then compared like coefficients to find that
m + n = 1 , m n = − 2 and 2 ( m − n ) = − 6 ⟹ m − n = − 3 .
I then added the first and final of these equation to get 2 m = − 2 ⟹ m = − 1 ,
which in turn gives us n = 2 . Fortunately this solution satisfied m n = − 2 , so I could conclude that the quartic could be factored in the way I had hoped it could. So I guess it was just a matter of making an educated guess, based on experience, on how the quartic could be factored and then going through the process to find the specific values necessary.
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Thank you! It really helped.
This equation can be rewritten as
x 4 + x 3 − 2 x 2 − 6 a x − 4 a 2 = 0 ⟹ ( x 2 − x − 2 a ) ( x 2 + 2 x + 2 a ) = 0
⟹ ( x − 2 1 − 1 + 8 a ) ( x − 2 1 + 1 + 8 a ) ( x + ( 1 + 1 − 2 a ) ) ( x + ( 1 − 1 − 2 a ) ) = 0 .
So in order to have real solutions we require that both
1 + 8 a ≥ 0 ⟹ a ≥ − 8 1 and 1 − 2 a ≥ 0 ⟹ a ≤ 2 1 .
Thus m + n = − 8 1 + 2 1 = 8 3 = 0 . 3 7 5 .