AMC 10A 2021 Problem 14 | Vieta's Formula

Try this beautiful Problem based on Vieta's Formula from AMC 10A, 2021 Problem 14.

Vieta's Formula | AMC 10A 2021, Problem 14

All the roots of the polynomial $z^{6}$-$10 z^{5}$+$A z^{4}$+$B z^{3}$+$C z^{2}$+$D z+16$ are positive integers, possibly repeated. What is the value of $B$ ?

Key Concepts

Vieta's Formula

Polynomial

Roots of the polynomial

Suggested Book | Source | Answer

Problem-Solving Strategies by Arthur Engel

AMC 10A 2021 Problem 14

-88

Try with Hints

Find out the degree of the given polynomial.

We know, Degree of polynomial= Number of roots of that polynomial.

Apply Vieta's Formula on the given polynomial.

By Vieta's Formula, the sum of the roots is 10 and product of the roots is 16.

Since there are 6 roots for this polynomial. By trial and check method find the roots.

The roots should be $2, 2, 2, 2, 1, 1$.

Now using the roots reconstruct the polynomial.

So the polynomial should be -

$(z-1)^{2}(z-2)^{4}$

$=(z^{2}-2 z+1)\\(z^{4}-8 z^{3}+24 z^{2}-32 z+16)$

Now equate it with the given polynomial to find the value of $B.$

Watch The Solution

Other useful links

Related Program

AMC - AIME Program at Cheenta

Subscribe to Cheenta at Youtube

Test of Mathematics Solution Subjective 176 - Value of a Polynomial at x = n+1

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 176 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

Suppose that P(x) is a polynomial of degree n such that $ P(k) = \frac {k}{k+1} $ for k = 0, 1, 2, ..., n . Find the value of P(n+1).

Solution

Consider an auxiliary polynomial g(x) = (x+1)P(x) - x . g(x) is an n+1 degree polynomial (as P(x) is n degree and we multiply (x+1) with it). We note that g(0) = g(1) = ... = g(n) = 0  (as the given condition allows (k+1) P(k) - k = 0 for all k from 0 to n). Hence 0, 1, 2, ... , n are the n+1 roots of g(x).

Therefore we may write g(x) = (x+1)P(x) - x = C(x)(x-1)(x-2)...(x-n) where C is a constant. Put x = -1. We get g(-1) = (-1+1)P(-1) - (-1) = C(-1)(-1-1)(-1-2)...(-1-n).

Thus 1 = C $ (-1)^{(n+1) } (n+1)! $ gives us the value of C. We put the value of C in the equation (x+1)P(x) - x = C(x)(x-1)(x-2)...(x-n) and replace x by n+1 to get the value of P(n+1).

$ (n+2)P(n+1) - (n+1) = \frac { (-1)^{(n+1)}}{(n+1)!} (n+1)(n)(n-1) ... (1) $ implying $ P(n+1) = \frac { (-1)^{(n+1)} + (n+1)}{(n+2)} $

Test of Mathematics Solution Subjective 175 - Integer Roots

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 175 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

 Let \(\text{P(x)}=x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\dots+a_{1}x+a_{0}\) be a polynomial with integer coefficients,such that,\(\text{P(0) and P(1)}\)  are odd integers.Show that

(a)\(\text{P(x)}\) does not have any even integer roots. 

(b)\(\text{P(x)}\) does not have any odd integer roots.

Solution

Given the two statements (a) and (b)  above it is clear that if we can prove that \(\text{P(x)}\) has no integer roots,then we are done.

 Let us assume \(\text{P(x)}\) has an integer root \(\text{ a}\).

Then we can write,

$$ \text{P(x)=(x-a)Q(x)}\dots(1)$$

where,\(\text{Q}\)is any function of \(\text{ x}\).

Now ,putting \(\text{x=0,1}\) in \(\text{ 1}\),we get

$$\text{P(0)=(-a)Q(0)}\dots(2)$$

and,

$$\text{P(1)=(1-a)Q(1)}\dots(3)$$

now as \(\text{(1-a)}\) and \(\ \text{(-a)}\),are consecutive integers \( \text{(2)}\) and \(\text{(3)}\)

cannot be both odd,

which means that \(\text{P(0),P(1)}\) cannot be both odd,given whatever \(\text{Q(0) and Q(1)}\) are.

So,it is a contradiction!!.

So,we can conclude that there exists no integer solution of \(\text{P(x)}\).

Hence,we are done.

Test of Mathematics Solution Subjective 75

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 75 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

Show that there is at least one value of $ {x}$ for which ${\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} = 1}$

Solution

$ {\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} = 1}$
$ {\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} - 1}$ is a continuous function. Now if we can chose $ {\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} - 1}$ takes both negative & positive numbers then the $ {\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} - 1 = 0}$ have a solution.

For $ {\displaystyle{x = {\frac{1}{64}}}}$, $ {\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} - 1 < 0}$ & for $ {x = 64}$, $ {\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} - 1 > 0}$.
So by intermediate value theorem we can say that $ {\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} - 1 = 0}$ has a solution for some real $ {x}$.

Watch Video Solution

Important Resources:

Test of Mathematics Solution Subjective 72 - Polynomial Problem

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 72 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

If $ {\displaystyle{\alpha}, {\beta}, {\gamma}} $ are the roots of the equation $ {\displaystyle{x^3 + px^2 + qx + r = 0}} $, find the equation whose roots are $ {\displaystyle{\alpha} - {\frac{1}{{\beta}{\gamma}}}} $, $ {\displaystyle{\beta} - {\frac{1}{{\alpha}{\gamma}}}} $, $ {\displaystyle{\gamma} - {\frac{1}{{\alpha}{\beta}}}} $ .

Solution

${\displaystyle{\alpha}, {\beta}, {\gamma}} $ are roots of $ {\displaystyle{x^3 + px^2 + qx + r = 0}} $
${\Rightarrow} $ $ {\displaystyle{\alpha}, {\beta}, {\gamma}} $ = $ {- r} $,
$ {\alpha + \beta + \gamma} $ = $ {- p} $
$ {{\alpha}{\beta} + {\gamma}{\alpha} + {\beta}{\gamma}} $ = $ {q} $
Now,
$ {\displaystyle{\left({\alpha} - {\frac{1}{{\beta}{\gamma}}}\right)}} $ + $ {\displaystyle{\left({\beta} - {\frac{1}{{\alpha}{\gamma}}}\right)}} $ + $ {\displaystyle{\left({\gamma} - {\frac{1}{{\alpha}{\beta}}}\right)}} $

= $ {\displaystyle{\frac{{\alpha}{\beta}{\gamma} - 1}{{\alpha}{\beta}{\gamma}}}} $ $ {\displaystyle{\left({\alpha} + {\beta} + {\gamma}\right)}} $

= - $ {\displaystyle{\frac{p(r+1)}{r}}} $

$ {\sum}$ $ {\displaystyle{\left({\alpha} - {\frac{1}{{\beta}{\gamma}}}\right)}} $ $ {\displaystyle{\left({\beta} - {\frac{1}{{\alpha}{\gamma}}}\right)}} $

= $ {\sum} $ $ {\displaystyle{\alpha \beta} - 2{\frac{1}{\gamma} + {\frac{1}{\alpha \beta \gamma^2}}}} $

= $ {\displaystyle{(\alpha \beta + \beta \gamma + \gamma \alpha)}} $ - 2$ {\displaystyle{\left({\frac{1}{\alpha}} + {\frac{1}{\beta}} + {\frac{1}{\gamma}}\right)}} $ + $ {\displaystyle{\frac{1}{\alpha \beta \gamma}}} $ + $ {\displaystyle{\left({\frac{1}{\alpha}}+{\frac{1}{\beta}} + {\frac{1}{\gamma}}\right)}} $

= $ {\displaystyle{q} - 2 {\left({\frac{q}{-r}}\right)} + {\frac{1}{-r}} {\left({\frac{q}{-r}}\right)}} $

= $ {\displaystyle{q}\left(1 + {\frac{1}{r}}\right)^2} $

$ {\displaystyle{\left({\alpha} - {\frac{1}{\beta \gamma}}\right)}} $ + $ {\displaystyle{\left({\beta} - {\frac{1}{\alpha \gamma}}\right)}} $ + $ {\displaystyle{\left({\gamma} - {\frac{1}{\alpha \beta}}\right)}} $

= $ {\displaystyle{\alpha \beta \gamma}} $ - 3 + $ {\displaystyle{3}{\frac{1}{\alpha \beta \gamma}}} $ - $ {\displaystyle \left({\frac{1}{\alpha \beta \gamma}}\right)^2} $

= - $ {\displaystyle \left(r + 3 + 3 {\frac{1}{r}} + {\frac{1}{r^2}}\right)} $

So the equation whose roots are ${\displaystyle{\alpha} - {\frac{1}{{\beta}{\gamma}}}} $, $ {\displaystyle{\beta} - {\frac{1}{{\alpha}{\gamma}}}} $, $ {\displaystyle{\gamma} - {\frac{1}{{\alpha}{\beta}}}} $ is ${\displaystyle{x^3} + {\frac{p(r+1)}{r}} x^2 + q \left(1+{\frac{1}{r}}\right)^2{x} + \left(r+3+{\frac{3}{r}}+{\frac{1}{r^2}}\right)} $

Test of Mathematics Solution Subjective 70 - Equal Roots

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 70 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

Suppose that all roots of the polynomial equation
$ {\displaystyle{x^4 - 4x^3 + ax^2 +bx + 1}} $ = 0 are positive real numbers.
Show that all the roots of the polynomial are equal.

Solution

$ {\displaystyle{x^4 - 4x^3 + ax^2 +bx + 1}} $ = 0
If the roots are $ {\displaystyle{\alpha}} $, $ {\displaystyle{\beta}} $, $ {\displaystyle{\gamma}} $ and $ {\displaystyle{\lambda}} $ .
then $ {\displaystyle{\alpha}} $, $ {\displaystyle{\beta}} $, $ {\displaystyle{\gamma}} $ and $ {\displaystyle{\lambda}} $ = 1
& $ {\displaystyle{\alpha}} $ + $ {\displaystyle{\beta}} $ + $ {\displaystyle{\gamma}} $ + $ {\displaystyle{\lambda}} $ = 4.
Now all of $ {\displaystyle{\alpha}} $, $ {\displaystyle{\beta}} $, $ {\displaystyle{\gamma}} $ and $ {\displaystyle{\lambda}} $ are positive so AM-GM inequality is applicable.
$ {\displaystyle{\frac{\alpha + \beta + \gamma + \lambda}{4}}}{\ge}$ $ {(\alpha\beta\lambda)^{\frac{1}{4}}}$
$ {\Rightarrow} $ $ {\frac{4}{4}} {\ge}$ $ {1^{\frac{1}{4}}}$
$ {\Rightarrow} $ 1 $ {\ge} $ 1
Now we know equality in AM-GM occours if all the numbers are equal.So $ {\displaystyle{\alpha}} $, $ {\displaystyle{\beta}} $, $ {\displaystyle{\gamma}} $ and $ {\displaystyle{\lambda}} $ are all equal.

Test of Mathematics Solution Subjective 69 - Coefficients of Polynomial

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 69 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

Suppose that the three equations $ {ax^2} - {2bx + c = 0} $ , \( {bx^2} –  {2cx + a = 0} \), and $ {cx^2} $ – $ {2ax + b = 0} $ all have only positive roots. Show that \( a =b = c\).

Solution

If possible let \(a\), \(b\), \(c\) are not all equal \( {ax^2}  – {2bx + c = 0} \) , \( {bx^2}  –  {2cx + a = 0}\), \({cx^2} - {2ax + b = 0}\) all have only positive roots. So all of \(a\), \(b\), \(c\) cannot be its same sign as discriminant > 0 for three equations ( \({b^2} >  {ac}\), \({a^2}  > {bc}\), \({c^2} >{ab}\) ). Without loss of generality we can assume \( a > b > c \) or \(a > c > b\) as the equations are cyclic. Now we know, \(\frac{\pm{\sqrt{b^2-ac}}}{a} > 0 \), \(\frac{\pm{\sqrt{a^2-bc}}}{c} > 0\), \(\frac{\pm{\sqrt{c^2-ab}}}{b} > 0 \). Now there 2 possibilities either b, c both are positive or one of b, c is positive. If b, c both are positive then,\( \frac{b-{\sqrt{b^2-ac}}}{a} < 0 \) [ not possible ]. If one of b, c is positive then, \( \frac{c-{\sqrt{c^2-ab}}}{b} < 0 \) [ not possible ] So a, b, c have to be equal.

Test of Mathematics Solution Subjective 66 - Range of a Polynomial

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 66 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

If c is a real number with 0 < c < 1, then show that the values taken by the function y = $ {frac {x^2+2x+c}{x^2+4x+3c}} $ , as x varies over real numbers, range over all real numbers.

Solution

y = $ {\frac {x^2+2x+c}{x^2+4x+3c}} $
or $ {yx^2} $ + 4yx + 3cy = $ {x^2} $ + 2x + c
or $ {x^2} $ (y-1) + (4y-2)x + 3cy - c = 0
Now if we can show that the discriminant $ {\ge{0}} $ for all y then for all real y there exist a real x.
Now, discriminant is $ {(4y-2)^2} $ - 4(y-1)(3cy-c)
We need to show $ {(4y-2)^2} $ - 4(y-1)(3cy-c) > 0 for any 0 < c < 1. (16-12c) $ {4^2} $ - (16-16c)y + (4-4c) > 0.
This is parabola opening upword.
Now if its discriminant < 0 then this equation is always > 0.
So this is equivalent to prove
$ {(16-16c)^2} $ - 4 (16-12c)(4-4c) < 0.
or $ {(2-2c)^2} $ - (4-3c)(1-c) < 0
or $ {c^2} $ -C < 0

Now given 1>c>0 so $ {c^2} $ <c

or $ {c^2} $ -c < 0

Conclusion: So for 1>c>0 y = $ {\frac{x^2+2x+c}{x^2+4x+3c}} $

Range over all real number when x varies over all real number.

Test of Mathematics Solution Subjective 61 - Symmetric Polynomial

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 61 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

Solve $ {{6x}^{2}} $ - 25x + 12 + $ {\frac{25}{x}} $ + $ {\frac{6}{x^2}} $ = 0.

Solution

$ {{6x}^{2}} $ – 25x + 12 + $ {\frac{25}{x}} $ + $ {\frac{6}{x^2}} $ = 0
= 6( $ {x^2} $ + $ {\frac{1}{x^2}} $) – 25(x - $ {\frac{1}{x}} $) +12 = 0
= 6 (x – $ {\frac{1}{x}})^2 $ – 25 (x – $ {\frac{1}{x}} $) + 24 = 0
= (x – $ {\frac{1}{x}} $) = $ {\frac{25\pm {\sqrt49}}{12}} $ = $ {\frac{3}{2}} $ or $ {\frac{8}{3}} $
If (x – $ {\frac{1}{x}} $) = $ {\frac{3}{2}} $
Or $ {{2x}^2} $ – 3x – 2 = 0
Or x = 2 , - $ {\frac{1}{2}} $
If (x – $ {\frac{1}{x}} $) = $ {\frac{8}{3}} $
Or $ {x^2} $ – 1 = $ {\frac{8}{3x}} $
Or $ {3x^2} $ – 8x – 3 = 0
Or x = 3, -1/3
X = 2,- $ {\frac{1}{2}} $ ,3,- $ {\frac{1}{3}} $

Using auxiliary polynomials in ISI Entrance

Auxiliary polynomials are useful for solving complicated polynomial problems. This problem from ISI Entrance (and a Soviet Olympiad) is a good example.

Watch video

Read More