What does a zero matter

Algebra Level 1

Two of the roots of the equation a x 3 + b x 2 + c x + d = 0 ax^3 + bx^2 + cx + d = 0 are 2 − 3 2 - \sqrt{3} and 5 5 , where a , b , c , d a,b,c,d are rational numbers. If the third one is m + n m + \sqrt{n} for some integers m m and n n , find m + n m + n


The answer is 5.

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Manish Mayank by Manish Mayank , 20, India

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4 solutions

Manish Mayank
May 28, 2015

In a polynomial equation with rational coefficients, irrational roots enter as conjugate pairs. So the third root will be 2 + 3 2 + \sqrt{3} . Hence the answer is 2 + 3 = 5 2+3=5

13 Helpful 0 Interesting 0 Brilliant 1 Confused

Moderator note:

Correct. Other than the three roots found: 2 + 3 , 2 − 3 , 5 2 +\sqrt3, 2- \sqrt3, 5 , what possible values of x x can satisfy this equation? If there is none, why not?

@Calvin Lin By L'Hospital's rule and the conditions of the problem, the sum and product of the roots of the equation are rational numbers. Thus, the third root is equal to k + 3 k+\sqrt{3} for some rational k . k. The product of the roots is 5 ( 2 − 3 ) ( k + 3 ) = 10 k − 5 k 3 + 10 3 − 15. 5(2-\sqrt{3})(k+\sqrt{3})=10k-5k\sqrt{3}+10\sqrt{3}-15. This is a rational number; the only way the (irrational) terms with 3 \sqrt{3} can be eliminated is if k = 2. k=2.

Trevor B. - 6 years ago

According to fundamental theorem of Algebra any polynomial eqn of degree n can have atmost n roots

sanghamitra ghosh - 3 years, 1 month ago
Akshay Yadav
Jun 2, 2015

A simple concept related to polynomial and their roots is related here. In the given polynomial, one of the root was integer i.e. 5 and the other irrational i.e. 2-(3)^0.5. Now if we the given polynomial which is of degree 3 with rational root then we'll find a polynomial with degree 2. Now this polynomial will have two roots out of which one is 2-(3)^0.5,then from our knowlegde we can tell that the other root will be conjungate of the previous i.e. with signs changed. So, the other root will be 2+(3)^0.5. And our required answer is 2+3=5.

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Rushi Bhatt
Jun 2, 2015

It has 4 roots one is 5 other is 2-3^1/2

so out of other 2 roots one would be conjugate of 2-3^1/2 so conjugate(2-(3^1/2)) will be 2+(3^1/2) so m is 2 and n is 3 so m+ n=5

0 Helpful 0 Interesting 0 Brilliant 0 Confused

How can a cubic polynomial have more than 3 zeroes

Manish Mayank - 6 years ago
Harshit Agarwal
May 29, 2015

In a polynomial equation with rational coefficients, irrational roots enter as conjugate pairs. So the third root will be 2+3^1/2 . Hence the answer is 2+3=5

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Right. Is there a unique solution for integer a a in the given equation?

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