Mathematics · Quantitative Aptitude

Polynomial and Quadratic Equations

239 Questions

Polynomial and quadratic equations involve finding the roots of expressions with varying degrees. The focus is on the relationship between coefficients and roots, symmetric functions, and solving higher degree polynomials. This topic is a core component of algebra in competitive testing.

Roots of quadratic equationsSymmetric root functionsCubic polynomialsRoot approximation methodsSum and product of roots

Polynomial and Quadratic Equations Questions

Multiple choice the nth roots of unity complex numbers maths

If $\alpha $ is a non-real root of $x^6=1$, then $\displaystyle \frac{\alpha ^5+\alpha ^3+\alpha +1}{\alpha ^2+1}=$

  1. $\alpha ^2$
  2. $0$
  3. $-\alpha ^2$
  4. $\alpha $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation


$1+\alpha+...+\alpha^{5}=0$ [sum of n roots of unity]
$\Rightarrow 1+\alpha +{ \alpha  }^{ 3 }+{ \alpha  }^{ 5 }=-\left( { \alpha  }^{ 2 }+{ \alpha  }^{ 4 } \right) $
$\displaystyle \Rightarrow 1+\alpha +{ \alpha  }^{ 3 }+{ \alpha  }^{ 5 }=-{ \alpha  }^{ 2 }\left( { 1+\alpha  }^{ 2 } \right) $
$\displaystyle \Rightarrow \frac { 1+\alpha +{ \alpha  }^{ 3 }+{ \alpha  }^{ 5 } }{ { 1+\alpha  }^{ 2 } } =-{ \alpha  }^{ 2 }$

Multiple choice the nth roots of unity complex numbers maths

If $(2 + i \sqrt 3)$ is a root of the equation $x^2 + px + q = 0$, where p and q are real, then (p, q) equals to

  1. $(4, 7)$
  2. $(-4, -7)$
  3. $(-4, 7)$
  4. $(4, -7)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since $2 + i \sqrt 3$ is one root, then other root will be $2 - i \sqrt 3$.
$\therefore x^2 + px + q = 0$ is given equatiion
$\therefore$ Sum of roots $= 2 + i \sqrt 3 + 2 - i \sqrt 3 = p$
$\therefore p = - 4$
Product of roots $q = 4 + 3 = 7$

Multiple choice the nth roots of unity complex numbers maths

If $2 + i$ and $\sqrt {5} - 2i$ are the roots of the equation $(x^{2} + ax + b)(x^{2} + cx + d) = 0$, where $a, b, c, d$ are real constants, then product of all roots of the equation is

  1. $40$
  2. $9\sqrt {5}$
  3. $45$
  4. $35$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$2 - i$ and $\sqrt {5} + 2i$ are other roots.
So, Product is $(2 + i)(2 - i)(\sqrt {5} + 2i)(\sqrt {5} - 2i)$
$= 5\times 9 = 45$

Multiple choice the nth roots of unity complex numbers maths

If $ 1,\alpha ,\alpha ^{2} .....\alpha ^{n-1}$ are n roots of unity then ,$1.\alpha .\alpha ^{2}....\alpha ^{n-1}$ equals

  1. $\left ( -1 \right )^{n-1}$

  2. 0

  3. 1

  4. -1
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$1,\alpha ,{ \alpha  }^{ 2 }...{ \alpha  }^{ n-1 }$ are the nth roots of unity. Thus, they are solutions of the equation: ${ x }^{ n }-1=0$. 
Thus, product of roots $= { (-1) }^{ n }(\dfrac { -1 }{ 1 } )$
(i.e. ${ (-1) }^{ n }$constant term / coefficient of ${ x }^{ n }$)
Thus, the product = ${ (-1) }^{ n }(\dfrac { -1 }{ 1 } )={ (-1) }^{ n }(\dfrac { -1 }{ 1 } )={ (-1) }^{ n+1 }={ (-1) }^{ 2 }{ (-1) }^{ n-1 }=1{ (-1) }^{ n-1 }={ (-1) }^{ n-1 }$
Hence, (A) is correct.

Multiple choice reciprocal equations theory of equations maths

The equation $3x^4-5x^3+3x^2-4x+5=0$ is of the type

  1. Quadratic

  2. Linear

  3. Reiprocal

  4. None of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given equation is $3x^4-5x^3+3x^2-4x+5=0$ .... $(i)$

The maximum power of $x$ in this equation is $ 4$, so this is $4th$ degree equation  
So, $(i)$ is neither linear nor quadratic equation.
In reciprocal equation ($ ax^4 +bx^3 +cx^2 +dx +e = 0 $)  the multiplication of the roots i.e ($\dfrac{e}{a}$) should be  $1$

In the given equation $(i)$, $\dfrac{e}{a}  = \dfrac{5}{3}$  
So the multiplication of roots is not equal to one
Therefore, equation $(i)$ is not a reciprocal equation.

Hence, option D is correct.

Multiple choice reciprocal equations theory of equations maths

The equation $2x^4-9x^3+14x^2-9x+2=0$ is of the type

  1. Quadratic equation

  2. Linear equation

  3. Reciprocal Equation

  4. None
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given equation is $2x^4-9x^3+14x^2-9x+2=0$ .... $(i)$

The maximum power of $x$ in this equation is $ 4$, so this is $4th$ degree equation  
So, it is neither linear nor quadratic equation.
In reciprocal equation $ ax^4 +bx^3 +cx^2 +dx +e = 0 $, the multiplication of the roots i.e ($\dfrac{e}{a}$) should be  $1$

In the given equation $(i)$, $\dfrac{e}{a}  = \dfrac{2}{2}$  
So the multiplication of roots is equal to one
Therefore, equation $(i)$ is a reciprocal equation.

Hence, option C is correct.

Multiple choice reciprocal equations theory of equations maths

What is a reciprocal equation?

  1. It involves reciprocal of the given variable.

  2. It involves square of the given variable.

  3. It involves squareroot of the given variable.

  4. It involves square and reciprocal of the given variable.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Reciprocal is said to be divide $1$ by a number. Reciprocal equation involves reciprocal of the given number.

Multiple choice reciprocal equations theory of equations maths

Determine the root of the equation: $\dfrac{9}{x}-\dfrac{7}{x}=1$

  1. $x=2$
  2. $x=-2$
  3. $x=1$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given reciprocal equation can be written as

$\dfrac{9}{x}=\dfrac{7+x}{x}$
Cancelling out the denominator on both side, we get
$9=7+x$
$\Rightarrow x=2$
Hence, option A is correct.

Multiple choice reciprocal equations theory of equations maths

If $b$ is a root of a reciprocal equation, $f(x)=0$, then another root of $f(x)=0$ is:

  1. $\dfrac{-1}{b}$
  2. $\dfrac{1}{b^2}$
  3. $\sqrt b$
  4. $\dfrac{1}{b}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A reciprocal equation is an equation whose roots can be divided into pairs of numbers, each the reciprocal of the other. 

(equivalently) an equation which is unchanged if the variable $x$ is replaced by its reciprocal $\dfrac{1}{x}$ is reciprocal equation.
Since one root is $b$, then the other root is $\dfrac{1}{b}$

Hence, option D is correct.

Multiple choice reciprocal equations theory of equations maths

A ............ equation is one which remains the same when $x$ is replaced by $\dfrac{1}{x}$.

  1. Reciprocal equation

  2. Radical equation

  3. Exponential equation

  4. Linear equation
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

(Originally) an equation whose roots can be divided into pairs of numbers, each the reciprocal of the other is called Reciprocal equation

(Equivalently) an equation which is unchanged if the variable $x$ is replaced by its reciprocal $\dfrac{1}{x}$ is known as reciprocal equation.
Hence, option A is correct.

Multiple choice reciprocal equations theory of equations maths

The roots of equation $2x^4-9x^3+14x^2-9x+2=0$ are

  1. $(1,2,3,4)$
  2. $\left(1,1,\dfrac{1}{2},2\right)$
  3. $\left(1,\dfrac{1}{3},3,1\right)$
  4. $\left(0,1,1,\dfrac{1}{2}\right)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We can see that in the giving equation, the multiplication of roots is $1$ 

i.e multiplication of roots $=\dfrac{e}{a}$ in the equation $ax^4 +bx^3 + cx^2 +dx  +e  =0$  
$\Rightarrow $ $\dfrac{e}{a}  = \dfrac{2}{2}  = 1 $
Now, sum of the roots is $\dfrac{-b}{a}  = -(\dfrac{-9}{2}) =  \dfrac{9}{2}$

$\Rightarrow $ it is only possible in option B.
Hence, option B is correct.

Multiple choice reciprocal equations theory of equations maths

Identify which of the following are reciprocal equations of 1st type.

  1. $2x^4+5x^3+2x^2+5x-2=0$
  2. $2x^4-5x^3+2x^2-5x+2=0$
  3. $2x^4-5x^3+2x^2+5x-2=0$

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Reciprocal equation is the equation which have even numbers of roots and if one root is $x$ then the other root will be $\dfrac{1}{x}$ and the multiplication of all roots will be one.

Now $1st$ type = where cofficients  $a=e$ in 4th order equation  $ax^4 +bx^3 +cx^{2}  + dx+e = 0 $

In option [A]  $a =2$  and $e = -2$  not $1st$ type

In option [B]  $a =2$  and $e = 2$  this is a $1st$ type reciprocal equation 

In option [C]  $a =2$  and $e = -2$  not a $1st$ type reciprocal equation.
Hence, B is correct.

Multiple choice reciprocal equations theory of equations maths

Identify if the following equation is a reciprocal equation by rearranging.

  1. $2(x^4+1)+89x^2= 56x(x^2+1)$
  2. $2(x^4+1)+89x^2= 56x(x+1)$
  3. $2(x^4+1)+89x^2= 56x^2(x+1)$

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

To be reciprocal equation, the multiplication of the roots $(\dfrac{e}{a})$ should be $1$ 

(A)
After rearranging the equation 
$\Rightarrow $   $2x^4 -56x^3 +89x^2 -56x +2 = 0 $
$\Rightarrow $  $\dfrac{e}{a} = \dfrac{2}{2}  = 1 $
So, multiplication of roots is $1$  
Thus, it is an reciprocal equation 

(B)
$2x^4 -33x^2 -56x +2 = 0 $

$\Rightarrow $  $\dfrac{e}{a} = \dfrac{2}{2}  = 1 $

So, multiplication of roots is $1$  
Thus, it is an reciprocal equation 


(C)

$\Rightarrow $   $2x^4 -56x^3 +33x^2 +2 = 0 $

$\Rightarrow $  $\dfrac{e}{a} = \dfrac{2}{2}  = 1 $

So, multiplication of roots is $1$  
Thus, it is an reciprocal equation.

Hence, the answer is option D.

Multiple choice reciprocal equations theory of equations maths

$2x^4-3x^3+7x^2+3x-2=0$ is not a reciprocal equation, because

  1. The coefficients from beginning to end and vice versa are not the same.

  2. All the coefficients of terms are not same

  3. The coefficients from beginning to end and vice versa are same.

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Here, the coefficients are not palindromic because the first and and last coefficients are opposite in sign , same is the case for second last and second coefficient.

Multiple choice reciprocal equations theory of equations maths

The Equation $5x^4-3x^3+7x^2-4x+2=0$ is of the type

  1. Quadratic

  2. Linear

  3. Reciprocal

  4. None
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The highest power of $x $ in this equation is $4$, so this is a $4th$ order equation.

Thus, it is neither linear nor quadratic.
Now to be reciprocal equation the multiplication of roots should be $1$  and in the given equation 
Multiplication of roots is $\dfrac{e}{a}=\dfrac{2}{5}$
So this not a reciprocal equation 
Hence, option D is correct.