Ugly Roots, Nice Answer

Algebra Level 3

Let the roots of the polynomial 2 x 3 − 3 x 2 + 5 x + c 2x^{3}-3x^{2}+5x+c (where c c is a constant) be α \alpha , β \beta , and γ \gamma . If 1 α β + 1 α γ + 1 β γ = α β + α γ + β γ , \frac{1}{\alpha\beta}+\frac{1}{\alpha\gamma}+\frac{1}{\beta\gamma}=\alpha\beta+\alpha\gamma+\beta\gamma, then the value of c c can be written as − m n -\frac{m}{n} , where m m and n n are positive, coprime integers. Find the value of m + n m+n .


The answer is 11.

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Walker Anderson by Walker Anderson , 20, USA

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2 solutions

From 1 α β + 1 β γ + 1 γ α = α β + β γ + γ α \displaystyle \frac{1}{\alpha\beta}+\frac{1}{\beta\gamma}+\frac{1}{\gamma\alpha} = \alpha\beta+\beta\gamma+\gamma\alpha

α + β + γ α β γ = α β + β γ + γ α \displaystyle \frac{\alpha+\beta+\gamma}{\alpha\beta\gamma} = \alpha\beta+\beta\gamma+\gamma\alpha

By Vieta's root formula;

3 2 − c 2 = 5 2 \displaystyle \frac{\displaystyle \frac{3}{2}}{\displaystyle \frac{-c}{2}} = \frac{5}{2}

c = − 6 5 \displaystyle c = \boxed{\displaystyle -\frac{6}{5}} ~~~

10 Helpful 0 Interesting 0 Brilliant 0 Confused

very easy solution , i also did the same

Raunak Agrawal - 6 years, 10 months ago

What on earth is Vieta's formula?

William Isoroku - 6 years, 10 months ago

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There you go Vieta's formula

Samuraiwarm Tsunayoshi - 6 years, 10 months ago

It is used to find the product or sum of the roots without actually finding its solution.

Here's how to derive Vieta's Formula for quadratic equations,

a ( x − α ) ( x − β ) = a x 2 − a ( α + β ) x + a ( α β ) a(x-\alpha)(x-\beta)=ax^2-a(\alpha+\beta)x+a(\alpha\beta)

A quadratic equation can be written as

a x 2 + b x + c ax^2+bx+c

Therefore, by comparing, you can easily see that b a = − ( α + β ) \frac{b}{a}=-(\alpha+\beta) is the product of the roots

And the sum of the roots is c a = α β \frac{c}{a}=\alpha\beta

If you want to apply this to equations of higher degree, you can expand it and compare the variable too, like what I did.

Daniel Lim - 6 years, 10 months ago

I did the same thing :)

Romeo Gomez - 6 years, 10 months ago

does not deserve so many points

chirag shetty - 6 years, 8 months ago

same way!.Nice solution there. :)

Soumo Mukherjee - 6 years, 7 months ago

Please excuse the uppercase gamma, I couldn't find a lowercase gamma!

From Vietè's Formulae , we can get the relation between the coefficients of this polynomial.

The general cubic is of the form P x 3 + Q x 2 + R x + S Px^{3} + Qx^{2} + Rx + S Now Vietè gives us,

α + β + γ = − Q P \alpha + \beta + \gamma = -\frac{Q}{P}

α β + β γ + γ α = R P \alpha\beta + \beta\gamma + \gamma\alpha = \frac{R}{P}

α β γ = − S P \alpha\beta\gamma = -\frac{S}{P}

Now comparing the general cubic to our polynomial, we get

P = 2 P = 2

Q = − 3 Q = -3

R = 5 R = 5

S = c S = c

Now, we can further simplify the given condition. On simplification, we get :

α + β + γ α β γ = α β + β γ + γ α \frac{\alpha+\beta+\gamma}{\alpha\beta\gamma} = \alpha\beta + \beta\gamma + \gamma\alpha

Substituting the values of P,Q,R and S, we get :

− Q P − S P = R P \frac{-\frac{Q}{P}}{-\frac{S}{P}} = \frac{R}{P}

Q S = R P \frac{Q}{S} = \frac{R}{P}

− 3 c = 5 2 \frac{-3}{c} = \frac{5}{2}

Which gives c = − 6 5 = − m n -\frac{6}{5} = -\frac{m}{n}

Hence, m + n = 6 + 5 = 11 m+n = 6 + 5 = \boxed{11}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

\gamma gives γ \gamma

Samuraiwarm Tsunayoshi - 6 years, 10 months ago

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