Relating Roots to Coefficients in a Quartic

Algebra Level 3

If three of the roots of x 4 + A x 2 + B x + C = 0 x^4+Ax^2+Bx+C=0 are − 1 -1 , 2 2 , and 3 3 , then find 2 C − A B 2C-AB .


The answer is 198.

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Finn Hulse by Finn Hulse , 21, USA

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4 solutions

Kanthi Deep
May 11, 2014

Let the 4th root is' a 'now the sum roots is -1+2+3+a=0 (because the coeff of second highest power is zero) then we get 4th root is -4 Now findd The polynomial equation and by comparing with given equation We get A=-15,B=10,C=24 and now 2C-AB=198

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Sai Ram
Jul 28, 2015

f ( 1 ) = A − B + C = − 1. f(1)=A-B+C=-1.

f ( 2 ) = 4 A + 2 B + C = − 16. f(2)=4A+2B+C=-16.

f ( 3 ) = 9 A + 3 B + c = − 81. f(3)=9A+3B+c=-81.

Hence , A = − 15 , B = 10 , C = 24. A=-15 , B= 10 , C= 24.

Therefore 2 C − A B = 2 ( 24 ) − ( − 15 ) ( 10 ) = 48 + 150 = 198 . 2C-AB = 2(24)-(-15)(10)=48+150 = \boxed{198}.

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Thanh Viet
May 8, 2014

we have: A-B+C=-1; 4A+2B+C=-16; 9A+3B+C=-81; Hence, A=-15; B=10;C=24. So 2C-AB=198

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Can you show how you derived these equations? Otherwise, great job! :D

Finn Hulse - 7 years, 1 month ago

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f ( 1 ) = A + B + C = − 1 f(1)=A+B+C=-1

f ( 2 ) = 4 A + 2 B + C = − 16 f(2)=4A+2B+C=-16

f ( 3 ) = 9 A + 3 B + C = − 81 f(3)=9A+3B+C=-81

Solving above eqns, we get

A = − 15 , B = 10 , C = 24 \boxed{A=-15}, \boxed{B=10}, \boxed{C=24}

⇒ 2 C − A B = 198 \Rightarrow 2C-AB= \boxed{198}

Aneesh Kundu - 7 years ago

I'll have to say, I'm a bit disappointed at you on this one; this is the exact same question as a Mu Alpha Theta question in 1992. It would be nice if you could give credit to them by putting a Source: MAO 1992 on the bottom.

Daniel Liu - 7 years ago

Of course :) Since A-B+C=-1; 4A+2B+C=-16 then 3A+3B=-15 => A+B=-5. (1) Since A-B+C=-1; 9A+3B+C=-81 then 8A+4B=-80 => 2A+B=-20 (2) From (1) and (2) we have: A=-15. Then we can find B and C

Thanh Viet - 7 years, 1 month ago
Shaun Loong
May 11, 2014

In general, a quartic equation is in the form of x 4 − ( sum of roots ) x 3 + ( sum of product of 2 roots ) x 2 − ( sum of product of 3 roots ) x + ( product of all 4 roots ) = 0 x^{4}-(\text{sum of roots})x^{3}+(\text{sum of product of 2 roots})x^{2}-(\text{sum of product of 3 roots})x+(\text{product of all 4 roots})=0 . Since the coefficient of x 3 x^{3} is 0 0 , hence: − 1 + 2 + 3 + α = 0 ⇒ α = − 4 -1+2+3+\alpha=0 \Rightarrow \alpha=-4 Hence, we know the roots to the equation is − 4 , − 1 , 2 , 3 -4, -1, 2, 3 . The equation is: ( x + 4 ) ( x + 1 ) ( x − 2 ) ( x − 3 ) = 0 ⇒ x 4 − 15 x 2 + 10 x + 24 = 0 (x+4)(x+1)(x-2)(x-3)=0 \Rightarrow x^{4}-15x^{2}+10x+24=0

It remains to show that 2 C − A B = 2 ( 24 ) − ( − 15 ) ( 10 ) = 198 2C-AB=2(24)-(-15)(10)=\boxed{198} .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

u know i did it the same way but the first 2 times i got it wrong due to some calculational error, finally i got it correct in my third try. in all i lost around 50 points and gained 25. so in short today is a bad day for me

Aneesh Kundu - 7 years ago

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