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 finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

lf $a=\displaystyle \cos\frac{2\pi}{7}+i\sin\frac{2\pi}{7}, \alpha=a+a^{2}+a^{4}$ and $\beta=a^{3}+a^{5}+a^{6}$, then $\alpha, \beta$ are the roots of the equation

  1. $x^{2}+x+1=0$
  2. $x^{2}+x+2=0$
  3. $x^{2}+2x+2=0$
  4. $x^{2}+2x+3=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$a={ e }^{ i2\pi /7 }\ a^ 7=1\ \alpha +\beta $

$=a+a^ 2+a^ 3+a^ 4+a^ 5+a^ 6\ =\dfrac{a(a^ 6-1)}{(a-1)}\ =\dfrac{(a^ 7-a)}{(a-1)}\ =\dfrac{(1-a)}{(a-1)}$
$=-1$
$\alpha \beta =(a+a^ 2+a^ 4)(a^ 3+a^ 5+a^ 6)\ =(a^ 4+a^ 6+a^ 7+a^ 5+a^ 7+a^ 8+a^ 7+a^ 9+a^ {10})\ =(a^ 4+a^ 6+1+a^ 5+1+a+1+a^ 2+a^ 3)\ =(3+a+a^ 2+a^ 3+a^ 4+a^ 5+a^ 6)\ =(3-1)$
$=2 $
The equation can be written as
 $x^ 2-(\alpha +\beta )x +\alpha \beta  =x^ 2+x+2$
Hence, option B is correct.

Multiple choice maths calculations and mental strategies 1 equations from statements forming equations from statements writing mathematical statements

If a $\in { 1,2,3,4 } ,$ then number of equations of the form $x ^ { 2 } + a x + 1 = 0$ having real roots is

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For real roots, the discriminant D = a^2 - 4 >= 0. This implies a^2 >= 4. Given a in {1, 2, 3, 4}, the values satisfying this are a=2, 3, 4. There are 3 such values.

Multiple choice maths square root square root of perfect square finding square root of a number square root of a perfect square

If $a,b,c$ are three distinct positive real numbers then the number of real roots of $ax^2+2b|x|-c=0$ is

  1. $0$
  2. $2$
  3. $4$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

${ ax }^{ 2 }+2b{ |x| }-c=a{ |x| }^{ 2 }+2b|x|-c$

                       $|x| =\dfrac { -b\pm \sqrt { 4{ b }^{ 2 }+4ac }  }{ 2a }$
                             $=\dfrac { -2b\pm 2\sqrt { { b }^{ 2 }+ac }  }{ 2a }$
                             $=\dfrac { -b\pm \sqrt { { b }^{ 2 }+ac }  }{ a }$
                             $=\dfrac { -b+\sqrt { { b }^{ 2 }+ac }  }{ a }$ (|x| can't be negative)
$\therefore 2$ real roots                            

Multiple choice maths power and exponent power of powers laws of exponents and powers law of indices

The sum of roots of the equation $(1.25)^{1-x^2} = (0.4096)^{1+x}$

  1. Infinite
  2. $1$
  3. $2$
  4. $4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$\left ( 1.25 \right )^{1-x^{2}}=\left ( 0.4096 \right )^{1+x}$

$\left ( \dfrac{125}{100} \right )^{1-x^{2}}=\left ( \dfrac{4096}{10000} \right )^{1+x}$

$\left ( \dfrac{5}{4} \right )^{1-x^{2}}=\left ( \left ( \dfrac{8}{10} \right )^{4} \right )^{1+x}$

$\left ( \dfrac{5}{4} \right )^{1-x^{2}}=\left ( \dfrac{4}{5} \right )^{4+4x}$

$\left ( \dfrac{5}{4} \right )^{1-x^{2}}=\left ( \dfrac{5}{4} \right )^{-4-4x}$

$\Rightarrow 1-x^{2}=-4-4x$

$x^{2}-4x-5=0$

$x^{2}-5x+x-5=0$
$x(x-5)+1(x-5)=0$
$(x+1)(x-5)=0$
$x=-1,5$

Therefore, Sum of the roots of equation  is $4$
Multiple choice de moivre’s theorem and its applications demoivre's theorem complex numbers maths

Construct an equation whose roots are $n^{th}$ powers of the roots of the equation $\displaystyle x^{2}-2x\cos \theta +1= 0.$

  1. $\displaystyle x^{2}-2n\cos n\theta x+1= 0$
  2. $\displaystyle x^{2}-2n\cos \theta x+1= 0$
  3. $\displaystyle x^{2}-2\cos n\theta x+1= 0$
  4. $\displaystyle x^{2}-2\cos ^{n}\theta x+1= 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know, $\displaystyle \alpha = \cos \theta +i\sin \theta , \beta = \cos \theta -i\sin \theta $
$\displaystyle \alpha ^{n}= \cos n\theta +i\sin n\theta ,$
$\displaystyle \beta ^{n}= \cos n\theta -i\sin n\theta $
$\displaystyle S= 2\cos n\theta , P= 1 \therefore x^{2}-Sx+P= 0$
or $\displaystyle x^{2}-2\cos n\theta x+1= 0$ is the required equation.

Ans: C

Multiple choice de moivre’s theorem and its applications demoivre's theorem complex numbers maths

If $\alpha, \beta$ are the roots of the equation $u^2-2u+2=0$ and if $\cot\theta=x+1$, then $[(x+\alpha)^n-(x+\beta)^m]/[\alpha-\beta]$ is equal to

  1. $\displaystyle \frac {\sin n\theta}{\sin^n\theta}$
  2. $\displaystyle \frac {\cos n\theta}{\cos^n\theta}$
  3. $\displaystyle \frac {\sin n\theta}{\cos^n\theta}$
  4. $\displaystyle \frac {\cos n\theta}{\sin^n\theta}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

${ u }^{ 2 }-2u+2=0$
$\Longrightarrow \quad u=1\pm i$
So,$\alpha =1+i\quad and\quad \beta =1-i$
Now given that,$x=\cot { \theta  } -1$
so,$\displaystyle \frac { { (x+\alpha ) }^{ n }-{ (x+\beta ) }^{ n } }{ \alpha -\beta  } =\frac { { (\cot { \theta  } -1+1+i) }^{ n }-{ (\cot { \theta  } -1 }+1-i)^{ n } }{ 2i } \ \ $
$=\displaystyle \frac { { (\cot { \theta  } +i) }^{ n }-(\cot { \theta  } -i)^{ n } }{ 2i } =\frac { { (\cos { \theta  } +i\sin { \theta  } ) }^{ n }-{ (\cos { \theta  } -\sin { \theta  } ) }^{ n } }{ ({ \sin { \theta  }  })^{ n }(2i) } \ \ $
$=\displaystyle \frac { { e }^{ (in\theta ) }-{ e }^{ -(in\theta ) } }{ ({ \sin { \theta ) }  }^{ n }2i } \ \ $
=$\displaystyle \frac { (\cos { (n\theta ) } +i\sin { (n\theta )) } -(\cos { (n\theta ) } -i\sin { (n\theta )) }  }{ ({ \sin { \theta ) }  }^{ n }2i } =\frac { \sin { (n\theta ) }  }{ { (\sin { \theta ) }  }^{ n } } \ \ $

Multiple choice de moivre’s theorem and its applications demoivre's theorem complex numbers maths

If n is a natural number$ \ge$ 2, such that $z^n = (z+ 1)^n$, then 

  1. roots of equation lie on a straight line parallel to the y-axis

  2. roots of equation lie on a straight line parallel to the x-axis

  3. sum of the real parts of the roots is -[(n-1)/2]

  4. none of these
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

$\displaystyle { z }^{ n }={ \left( z+1 \right)  }^{ n }$   where, $n\ge 2$     ...(1)

$\displaystyle \Rightarrow { \left( \frac { z+1 }{ z }  \right)  }^{ n }=1=\cos { 0 } +i\sin { 0 } $

$\displaystyle \Rightarrow \frac { z+1 }{ z } ={ \left( \cos { 0 } +i\sin { 0 }  \right)  }^{ \frac { 1 }{ n }  }$

$\displaystyle \Rightarrow \frac { z+1 }{ z } =\cos { \frac { 2k\Pi  }{ n }  } +i\sin { \frac { 2k\Pi  }{ n }  } $      ...{De Moivre's Theorem}

$\displaystyle \Rightarrow z=\frac { -1 }{ 1-\cos { \frac { 2k\Pi  }{ n }  } -i\sin { \frac { 2k\Pi  }{ n }  }  } \quad =\frac { -1 }{ 2\sin { \frac { k\Pi  }{ n } \left( \sin { \frac { k\Pi  }{ n } -i } \cos { \frac { k\Pi  }{ n }  }  \right)  }  } =\frac { -1\left( \sin { \frac { k\Pi  }{ n } +i } \cos { \frac { k\Pi  }{ n }  }  \right)  }{ 2\sin { \frac { k\Pi  }{ n }  }  } $

$\displaystyle \Rightarrow z=\frac { -\left( 1+i\cot { \frac { k\Pi  }{ n }  }  \right)  }{ 2 } $

Where$ k=1,2,3,....,n-1 $     ..{Since at k=0, z is not defined}

$\because \quad Re\left( z \right) $ is constant.
Therfore roots of ${ z }^{ n }={ \left( z+1 \right)  }^{ n }$ lie on straight line parellel to y-axis.

$\displaystyle \because \quad z=\frac { -\left( 1+i\cot { \frac { k\Pi  }{ n }  }  \right)  }{ 2 } $ and $k=1,2,3,....,(n-1)$

Sum of $\displaystyle Re\left( z \right) $= $-\frac { \left( n-1 \right)  }{ 2 } $.

Ans: A,C

Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

If $a,b,c$ are distinct and the roots of $\left( b-c \right) { x }^{ 2 }+\left( c-a \right) x+\left( a-b \right) =0$ are equal, then $a,b,c $ are in

  1. Arithmetic progression

  2. Geometric progression

  3. Harmonic progression

  4. Arithmetico-Geometric progression
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Clearly $x=1$ is a solution
$\therefore$  product of the roots $=\dfrac { a-b }{ b-c }$ 
$\therefore \left( 1 \right) \left( 1 \right) =\dfrac { a-b }{ b-c }$ 
$\Longrightarrow b-c=a-b$
$\Longrightarrow2b=a+c\Longrightarrow a,b,c$ are in Arithmetic progression.

Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

If the roots of palynomial $P ( x ) = x ^ { 3 } - 3 x ^ { 2 } + k x + 4 $ are in $A P ,$ then $\left| k \right| $. Has the value equal to

  1. 0

  2. 1

  3. 2

  4. 3
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given roots of the polynomial are in AP

let the roots of the polynomial be $a-d,a,a+d$
$\quad a-d+a+a+d=-\frac { -3 }{ 1 } \ \Rightarrow 3a=3\ \Rightarrow a=1$
so, $a=1$ is one of the roots of the equation
$\quad p\left( 1 \right) ={ 1 }^{ 3 }-3\times { 1 }^{ 2 }+k+4=0\ \Rightarrow k+2=0\ \Rightarrow k=-2$
$\left| k \right| =2$

Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

If roots of the equation $(a-b)x^{2}+(c-a)x+(b-c)=0, a \neq b \neq c$ are equal, then $a,b,c$ are in 

  1. $A.P$
  2. $H.P$
  3. $G.P$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We have,

Given equation is

$\left( a-b \right){{x}^{2}}+\left( c-a \right)x+\left( b-c \right)=0$

On comparing that,

$A{{x}^{2}}+Bx+C=0$

Now,

$ A=\left( a-b \right) $

$ B=\left( c-a \right) $

$ C=\left( b-c \right) $

Roots are equal

Then,

$ D=0 $

$ {{B}^{2}}-4AC=0 $

$ \Rightarrow {{\left( c-a \right)}^{2}}-4\left( a-b \right)\left( b-c \right)=0 $

$ \Rightarrow {{c}^{2}}+{{a}^{2}}-2ac=4\left( ab-ac-{{b}^{2}}+bc \right) $

$ \Rightarrow {{c}^{2}}+{{a}^{2}}-2ac=4ab-4ac-4{{b}^{2}}+4bc $

$ \Rightarrow {{c}^{2}}+{{a}^{2}}-2ac+4ac=4ab-4{{b}^{2}}+4bc $

$ \Rightarrow {{\left( c+a \right)}^{2}}=4b\left( a-b+c \right) $

Hence, this is the answer

Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

If $a,b,c$ are distnct and the roots of $(b-c)x^{2}+(c-a)x+(a-b)=0 $are equal, then $a,b,c$ are in

  1. Arithmetic progression

  2. Geometric prograsson

  3. Harmonic prograssiion

  4. Arithmetco- Geometric prograssion
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a quadratic equation to have equal roots, its discriminant must be zero. Setting (c-a)^2 - 4(b-c)(a-b) = 0 leads to (c-a)^2 + 4(b-c)(b-a) = 0, which simplifies to (a+c-2b)^2 = 0, implying a+c = 2b, which is the definition of an arithmetic progression.