Equations and polynomials?
Let a , b , c , d , e be real numbers that satisfy the equations:
c + a = 1 5
a c + b + d = 8 5 ,
a d + b c + e = 2 2 5 ,
a e + b d = 2 7 4 ,
b e = 1 2 0 .
Find the sum of all possible values of e .
Here's a hint (a simpler problem)
This problem is part of the set ... and polynomials
The answer is 225.
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2 solutions
HOW TO FIND 5 REAL ROOTS? THEN HOW TO SELECT 3 ROOTS OUT OF FIVE FOR e? PLEASE ELABORATE THE METHOD.
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One solution is (x-x1) (x-x2) (x-x3)=(x^3+cx^2+dx+e),so e=-x1 x2 x3,other solutions are 1,2,4 in index of x, -x1 x2 x3-x1 x2 x4-x1 x2 x5-x2 x3 x4-...-x3 x4 x5=e1+e2+e3+...e10 is coefficient by x^2,and that is 225
What is this property called?
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I would call it "Using Vieta's formula".
It is similar to solving the system of equations
a
+
b
+
c
=
0
,
a
b
+
b
c
+
c
a
=
0
a
b
c
=
0
,
which has an obvious immediate application.
That's a nice approach :)
However, the quintic equation that you have given only has 1 real root. As such, we cannot take " x 3 + c x 2 + d x + e as the product of any three of these 5 factors". In order for the variables to be real, we must take the real root, along with a conjugate pair of complex roots.
As such, the problem, as currently stated, does not have a final answer of 225. Resorting to Wolfram Alpha to factorize the polynomial, the roots are α 1 = 6 . 9 7 8 5 3 , α 2 , α 3 = − 3 . 6 3 9 ± 3 . 1 0 1 0 3 i , α 4 , α 5 = − 0 . 3 7 1 7 3 1 ± 0 . 7 8 3 6 2 6 i .
Hence, this gives us an answer of α 1 α 2 α 3 + α 1 α 4 α 5 = − 1 6 4 . 7 6 9 8 .
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In his solution, Mr. Tan made a slight typo. He meant to write x 5 + 1 5 x 4 + 8 5 x 3 + 2 2 5 x 2 + 2 7 4 x + 1 2 0 = 0 . (If you look at the original problem, the coefficient of x should be 274, not 174.) This equation factors as ( x + 1 ) ( x + 2 ) ( x + 3 ) ( x + 4 ) ( x + 5 ) = 0 . I believe the original answer of 225 still stands.
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Thanks for pointing that out. We indeed have 5 real roots, and the conclusion follows. I have edited the solution accordingly.
I have updated the answer back to 225.
If we supose that a,b,c,d,e are positive integers then from b*e=120,must be that e|120. Number120 has 16divisors: 1,2,3,4,5,6,120,8,20,10,24,12,30,40,15,60.That are posible values for e,Than b=120/e and this system is 3x3,and need about 2minutes for solving. for e=1,2,3,4,5,120 there are no solutions and 6+8+10+12+15+20+24+30+40+60=225.
There is no reason to suppose that a, b, c, d, e must all be positive integers. You are only given that they are real numbers.
Can you also explain how you solved for e = 6 , 8 , 1 0 , … ? What would be the corresponding solutions? Note that we do not have a system of linear equations, due to the term a c in a c + b + d = 8 5 .
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For example e=10 => b=12 so our system becomes c+a=15 , ac+d=73 , ad+12c=215 , 5a+6d=137 so c=15-a,d=(137-5a)/6 so we get a(137-5a)/6+180-12a=215 i.e. a²-13a+42=0 so a=(13±√(169-168))/2=(13±1)/2 For a=7 we get c=8,d=17, and need just to check is it ac+d=73 ac+d=7*8+17=56+17=73 so for e=12 a=7,b=12,c=8,d=17 is solution. Same for other cases of e. This is another solution,we need to find prove why they must be positive integers not using factorization of polynomial. Now I see that a,b,c,d,e are integers because (x²+bx+c)(x^3+cx²+dx+e)=(x^5+15x^4+85x^3+225x²+274x+120)=(x+1)(x+2)(x+3)(x+4)(x+5)
I will leave a hint for those who still want to try:
What are the coefficients of the expansion of ( x 2 + a x + b ) ( x 3 + c x 2 + d x + e ) ?
If you don't want the solution, don't read further.
Solution: After expansion, compare coefficients to get the above expression equal to x 5 + 1 5 x 4 + 8 5 x 3 + 2 2 5 x 2 + 2 7 4 x + 1 2 0 = ( x − r 1 ) ( x − r 2 ) . . . ( x − r 5 ) where all r i , i = 1 , 2 , 3 , 4 , 5 are the roots. In fact the roots are all real. (Find them!)
This means ( x 3 + c x 2 + d x + e ) is the product of three of these 5 factors in the form ( x − r i ) . Hence e is the product of three of the numbers in the set of the negatives of the five roots. The sum of all such e , by Vieta's formulae, is just 225.