Cubic polynomials
Let f ( x ) , g ( x ) be two monic cubic polynomials and r a real number.
f ( x ) has roots r + 1 , r + 7 only.
g ( x ) has roots r + 3 , r + 9 only.
Suppose f ( x ) − g ( x ) = r for all x . Find r .
This problem is part of the set ... and polynomials
The answer is 32.
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3 solutions
You put the roots of f ( x ) in the place of g ( x ) . I think the answer is − 3 2 instead of 3 2 .
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no, the constant term is negative product of roots for odd degree polynomial, that's why we swap f and g
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Can you show me the polynomials f ( x ) and g ( x ) .
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@Sujoy Roy – I'll simplify it for you. f ( x ) = ( x − ( r + 1 ) ) ( x − ( r + 7 ) ) 2 and g ( x ) = ( x − ( r + 3 ) ) 2 ( x − ( r + 9 ) ) .
f ( x ) − g ( x ) = ( x − ( r + 1 ) ) ( x − ( r + 7 ) ) 2 − ( x − ( r + 3 ) ) 2 ( x − ( r + 9 ) )
Since it is = r for all x , Putting x = 0 , we get,
f ( 0 ) − g ( 0 ) = − ( r + 1 ) ( r + 7 ) 2 − ( − ( r + 3 ) 2 ( r + 9 ) )
⇒ r = ( r + 3 ) 2 ( r + 9 ) − ( r + 1 ) ( r + 7 ) 2 .
IMO, OP should have put x = r + 9 , it would've simplified the last expression.
@Sujoy Roy – np: f must be (x-33)(x-39)(x-39), g is (x-35)(x-35)(x-41)
Thats what I got too. But I realised my mistake.
Answer should be then...-32
In fact the answer will be the same even if the only word occurring at two places is dropped. Let the third roots be anything. Very good question.
I like how you generalized this problem.
Can you explain why we do not need the restriction on the third root?
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I will solve the problem allowing their third roots to be any real values.
For convenience, r1=(r+1), r3=(r+3), r7=(r+7) , r9=(r+9).
f(x) has roots, r1, r7 and s and g(x) has roots, r3, r9 and t.
So, for all x,
f(x) = (x-r1) (x-r7) (x-s) .................................. (1)
g(x) = (x-r3) (x-r9) (x-t) .................................. (2)
f(x) - g(x) = r ....................................................... (3) (given)
Substituting the roots of g(x), r3 and r9 , into x in (3) yields,
f(r3) = r and f(r9)=r ........................................(4a) and (4b)
Simplifying (4a):
(r3-r1) (r3-r7) (r3-s) = r
(2) (-4) (r + 3 - s) = r
9r - 8s = -24 ...........................................................(5a)
Simplifying (4b):
(r9-r1) (r9-r7) (r9-s) = r
(8) (2) (r + 9 - s) = r
15r - 16s = -144 ...................................................(5b)
Solving (5a) and (5b) yields
r = 3 2 and s=39.
Let the third root be (p + 2) for first equation and hence it is (p - 2) for the second, as sums r + 1 + r + 7 + p + 2 = r + 3 + r + 9 + p - 2 equal. Next (r+1)(r+7) + (p+2)(2r+8) = (r+3)(r+9) + (p - 2)(2r+12) and next products (r+3)(r+9)(p-2) - (r+1)(r+7)(p + 2) = r are the two equations that follow from the given statement f(x) - g(x) = r for all x. Eliminating p and solving for r gives the required answer.
f(x) and g(x) have same coefficients next to x²(because f(x)-g(x)=r) which means r+1+r+7+r+7-a=r+3+r+3+r+9+b (from Vieta's formulas)where a can be 0 or 6,and b can be 0 and 6 From this equation -b=a,so a=b=0 which means r+7 and r+3 are multiply roots.So we get f(x)-g(x)=-(r+1)(r+7)² +(r+3)²(r+9)=-49+81=32.So,r=32...
if roots of f(x) are r+1, r+7, r+a ( a = 1 or a = 7 ) and roots of g(x) are r+3, r+9, r+b ( b = 3 or b = 9 ) then (r+1) + (r+7) + (r+a) = (r+3) + (r+9) + (r+b) therefore a = 7, b = 3
next, r = f(x) - g(x) = (r+3)(r+3)(r+9) - (r+1)(r+7)(r+7) = 32