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## Roots of Equation and Vieta’s formula | AIME I, 1996 Problem 5

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1996 based on Roots of Equation and Vieta’s formula.

## Roots of Equation and Vieta’s formula – AIME I, 1996

Suppose that the roots of $x^{3}+3x^{2}+4x-11=0$ are a,b and c and that the roots of $x^{3}+rx^{2}+sx+t=0$ are a+b,b+c and c+a, find t.

• is 107
• is 23
• is 840
• cannot be determined from the given information

### Key Concepts

Functions

Roots of Equation

Vieta s formula

AIME I, 1996, Question 5

Polynomials by Barbeau

## Try with Hints

With Vieta s formula

$f(x)=x^{3}+3x^{2}+4x-11=(x-a)(x-b)(x-c)=0$

$\Rightarrow a+b+c=-3$, $ab+bc+ca=4$ and $abc=11$

Let a+b+c=-3=p

here t=-(a+b)(b+c)(c+a)

$\Rightarrow t=-(p-c)(p-a)(p-b)$

$\Rightarrow t=-f(p)=-f(-3)$

$t=-[(-3)^{3}+3(-3)^{2}+4(-3)-11]$

=23.

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## Roots of cubic equation | AMC-10A, 2010 | Problem 21

Try this beautiful problem from Algebra based Roots of the cubic equation.

## Roots of cubic equation – AMC-10A, 2010- Problem 21

The polynomial $x^3-ax^2+bx-2010$ has three positive integer roots. What is the smallest possible value of $a$?

• $98$
• $78$
• $83$
• $76$
• $90$

### Key Concepts

Algebra

Vieta’s Relation

roots of the equation

Answer: $78$

AMC-10A (2010) Problem 19

Pre College Mathematics

## Try with Hints

The given equation is $x^3-ax^2+bx-2010$.we have to find out the smallest possible value of $a$.From Vieta’s Relation we know that if $r_1,r_2,r_3$ are the roots of equation $ax^3+bx^2+cx+d=0$ then $r_1 +r_2+r_3= -\frac{b}{a}$ and $r_1 r_2 r_3=-\frac{d}{a}$

can you finish the problem……..

Therefore $a$ is the sum of the three roots of the polynomial $x^3-ax^2+bx-2010$. and $2010$  is the product of the three integer roots.Now the factors of $2010$ =$2 \times 3 \times 5 \times 67$.there are only three roots to the polynomial so out of four roots we have to choose three roots such that two of the four prime factors must be multiplied so that we are left with three roots. To minimize $a$, $2$ and $3$ should be multiplied,

can you finish the problem……..

Therefore the value of $a$=$6+5+67=78$

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## Cubical Box | AMC-10A, 2010 | Problem 20

Try this beautiful problem from Geometry based on cubical box.

## Cubical Box – AMC-10A, 2010- Problem 20

A fly trapped inside a cubical box with side length $1$ meter decides to relieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once. To get from a corner to any other corner, it will either fly or crawl in a straight line. What is the maximum possible length, in meters, of its path?

• $4\sqrt 5+3\sqrt 2$
• $4\sqrt 5+3\sqrt 2$
• $4\sqrt 3+4\sqrt 2$
• $7\sqrt 3+4\sqrt 2$
• $4\sqrt 3+7\sqrt 2$

### Key Concepts

Geometry

Cubical

Pythagoras

Answer: $4\sqrt 3+4\sqrt 2$

AMC-10A (2010) Problem 20

Pre College Mathematics

## Try with Hints

Given that “A fly trapped inside a cubical box with side length $1$ meter decides to relieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once”……………………………Therefore we may say that from a corner to any other corner the straight path will be $A \to G \to B \to H \to C \to E \to D \to F \to A$

can you finish the problem……..

The distance of an interior diagonal in this cube is $\sqrt 3$ ( i.e $HB$) and the distance of a diagonal on one of the square faces is $\sqrt 2$ ( i.e $HA$)

can you finish the problem……..

Now the fly visits each corner exactly once, it cannot traverse such a line segment twice. Also, the cube has exactly four such diagonals, so the path of the fly can contain at most four segments of length.Therefore  the maximum distance traveled is $4\sqrt 3+4\sqrt 2$