Too Many Values

Algebra Level 5

Find all rational numbers a a , b b and c c such that the equation x 3 + a x 2 + b x + c = 0 x^3 + ax^2+bx+c = 0 has roots a a , b b and c c .

Enter your answer as the sum of all solutions. For example, if ( a , b , c ) (a,b,c) are ( 5 , 6 , 7 ) (5,6,7) and ( 3 , 4 , 8 ) (3,4,8) , then enter your answer as 5 + 6 + 7 + 3 + 4 + 8 = 33 5+6+7+3+4+8=33 .

Note: b , c b,c cannot be simultaneously equal to 0 0 .


The answer is -2.

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Deepansh Jindal by Deepansh Jindal , 19, India

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2 solutions

Chew-Seong Cheong
Apr 19, 2016

Relevant wiki: Vieta's Formula Problem Solving - Basic

There are two cases: { c ≠ 0 : x 3 + a x 2 + b x + c = 0 c = 0 : x ( x 2 + a x + b ) = 0 \begin{cases} c \ne 0: & x^3+ax^2+bx+c = 0 \\ c = 0: & x(x^2+ax+b) = 0 \end{cases}

( Note: The question should mention that b and c cannot be = 0 simultaneously. \small \color{#D61F06}{(\text{Note: The question should mention that } b \text{ and } c \text{ cannot be = 0 simultaneously. }} Because when b = c = 0 , the remaining root is x = − a and the answer is − ∞ ) \small \color{#D61F06}{\text{Because when } b = c =0 \text{, the remaining root is }x=-a \text{ and the answer is } -\infty)}

Note that for c = 0 c=0 one of the roots is 0 0 which is equal to c c hence satisfying the conditions.

Case c ≠ 0 : x 3 + a x 2 + b x + c = 0 ‾ \underline{\text{Case }c\ne 0: \quad x^3+ax^2+bx+c = 0}

Using Vieta's formulas, we have:

{ a + b + c = − a . . . ( 1 ) a b + b c + c a = b . . . ( 2 ) a b c = − c . . . ( 3 ) \begin{cases} a + b + c = -a & ...(1) \\ ab+bc+ca = b & ...(2) \\ abc = -c &...(3) \end{cases}

\(\begin{array} {} (3): & abc = -c & \Rightarrow ab = -1 & \Rightarrow b = - \dfrac{1}{a} \\ (1): & a+b+c = -a & \Rightarrow a - \dfrac{1}{a} + c = -a & \Rightarrow c = \dfrac{1}{a} - 2a \end{array} \)

( 2 ) : a b + b c + c a = b a ( − 1 a ) + ( − 1 a ) ( 1 a − 2 a ) + ( 1 a − 2 a ) a = − 1 a Multiply both sides by − a a + ( 1 a − 2 a ) + ( 2 a 2 − 1 ) a = 1 Multiply both sides by a a 2 + 1 − 2 a 2 + 2 a 4 − a 2 = a Rearrange 2 a 4 − 2 a 2 − a + 1 = 0 ( a − 1 ) ( 2 a 3 + 2 a 2 − 1 ) = 0 \begin{aligned} (2): \quad ab+bc + ca & = b \\ a \left(-\frac{1}{a}\right) + \left(-\frac{1}{a}\right) \left(\frac{1}{a}-2a\right) + \left(\frac{1}{a}-2a \right)a & = -\frac{1}{a} \quad \quad \small \color{#3D99F6}{\text{Multiply both sides by }-a} \\ a + \left(\frac{1}{a}-2a\right) + \left(2a^2 - 1 \right)a & = 1 \quad \quad \small \color{#3D99F6}{\text{Multiply both sides by }a} \\ a^2 + 1 - 2a^2 + 2a^4 - a^2 & = a \quad \quad \small \color{#3D99F6}{\text{Rearrange}} \\ 2a^4 - 2a^2 - a + 1 & = 0 \\ (a-1)(2a^3+2a^2-1) & = 0 \end{aligned}

Using rational root theorem, we note that 2 a 3 + 2 a 2 − 1 2a^3+2a^2-1 has no rational root, therefore, the rational value of a a is only 1 1 .

Therefore, { a = 1 b = − 1 a = − 1 c = 1 a − 2 a = − 1 ⇒ ( a , b , c ) = ( 1 , − 1 , − 1 ) \begin{cases} a = 1 \\ b = - \dfrac{1}{a} = -1 \\ c = \dfrac{1}{a} - 2a = -1 \end{cases} \quad \Rightarrow (a, b, c) = (1,-1,-1)

Case c = 0 : x ( x 2 + a x + b ) = 0 ‾ \underline{\text{Case }c = 0: \quad x(x^2+ax+b) = 0}

{ a b = b ⇒ a = 1 a + b = − a ⇒ 1 + b = − 1 ⇒ b = − 2 ⇒ ( a , b , c ) = ( 1 , − 2 , 0 ) \begin{cases} ab = b & \Rightarrow a = 1 \\ a + b = -a & \Rightarrow 1 + b = -1 & \Rightarrow b = -2 \end{cases} \quad \Rightarrow (a, b, c) = (1,-2,0)

The required answer: 1 − 1 − 1 + 1 − 2 + 0 = − 2 1-1-1+1-2+0 = \boxed{-2}

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Deepansh Jindal , I have edited some of your problems. Please do not use all uppercase, it is equivalent to SHOOTING in writing, which is therefore RUDE. Learn up LaTeX \LaTeX . It is easy. You just need to close the text with a backslash "\" open brace "(" and backslash "\" close brace ")". Put your mouse cursor on top of the formulas to see the keystrokes.

Chew-Seong Cheong - 5 years, 1 month ago

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Thanks sir both for solution and modification in the question..

Deepansh Jindal - 5 years, 1 month ago
Aakash Khandelwal
Apr 20, 2016

We have only 3 solutions

( 0 , 0 , 0 ) , ( 1 , − 2 , 0 ) , ( 1 , − 1 , − 1 ) (0,0,0),(1,-2,0),(1,-1,-1)

0 Helpful 0 Interesting 0 Brilliant 0 Confused

@Aakash Khandelwal See @Chew-Seong Cheong solution , whenever b = c = 0 b = c = 0 , we have infinitely many solutions.

Ankit Kumar Jain - 4 years, 3 months ago

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