Mathematics · Quantitative Aptitude

Polynomial and Quadratic Equations

223 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 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.

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

Let $f(x)=3ax^{2}-4bx+c(a,b,c \in R, a \neq 0)$ where $a,b,c$ are in $A.P$. Then the equation $f(x)=0$ has

  1. No real solution.

  2. Two unequal real roots.

  3. Sum of roots always negative.

  4. Product of roots always positive.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since $a,b,c$ are in A.P., so,

$2b = a + c$

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

The discriminant of the given function$f\left( x \right) = 3a{x^2} - 4bx + c$ is,

$D = 16{b^2} - 12ac$

$ = 4{\left( {a + c} \right)^2} - 12ac$

$ = 4\left[ {\left( {{a^2} + {c^2} + 2ac} \right) - 3ac} \right]$

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

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

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

Case 1: If $a$ and$c$ are of opposite signs, then, $D = \left(  +  \right){\rm{ve}}$.

Case 2: If $a$ and$c$ are of same signs, then, $D = \left(  +  \right){\rm{ve}}$.

This shows that $f\left( x \right) = 0$ has two unequal real roots.

Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

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

  1. G.P.

  2. H.P.

  3. A.P.

  4. A.G.P.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$(b-c)x^{2}+(c-a)x+a-b=0$
Root are equal, so $D=0$
$\Rightarrow (c-a)^{2}-4(b-c)(a-b)=0$
$\Rightarrow c^2+a^2-2ac-4ab+4b^2+4ac-4bc=0$
$\Rightarrow c^2+a^2+4b^2+2ac-4ab-4bc=0$
$\Rightarrow (a+c-2b)^{2}=0$
$a+c=2b$

























Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

If the roots of ${ x }^{ 2 }-k{ x }^{ 2 }+14x-8=0$ are in geometric progression, then $k=$

  1. $-3$
  2. $7$
  3. $4$
  4. $0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\Rightarrow \frac { a }{ r } ,a,ar=8\quad \quad \Rightarrow { a }^{ 3 }=8\quad \quad \Rightarrow a=2$
$a=2$ is a root of the given equation 
$\Rightarrow 8-4k+28-8=0\quad \quad \Rightarrow k=7$

Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

If $\alpha, \beta, \gamma$ are non-constant terms in G.P and equations $\alpha { x }^{ 2 }+2\beta x+\gamma =0\quad $ and ${x}^{2}+x-1=0$ has a common root then $\left( \gamma -\alpha  \right) ,\beta $ is

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

Let the common ratio of G.P is $r$ Therefore $\quad \beta =\alpha t,\alpha { t }^{ 2 }$
Equation $\alpha { x }^{ 2 }+2\alpha rx+\alpha { t }=0\quad 
\Rightarrow { x }^{ 2 }+2rx+{ t }^{ 2 }=0....(i)$
Given equation (i) and ${ x }^{ 2 }+x-1=0....(ii)$ has a common root
$(i)-(ii)\Rightarrow (2e-1)x+({ r }^{ 2 }+1)=0\Rightarrow x=\cfrac { -\left( { r }^{ 2 }+1 \right)  }{ 2r-1 } ....(iii)\quad $
Putting (iii) in equation (ii) $\Rightarrow { \left( { r }^{ 2 }+1 \right)  }^{ 2 }-\left( { r }^{ 2 }+1 \right) (2r-1)-{ \left( { 2r }^{ 2 }-1 \right)  }^{ 2 }=0\Rightarrow { r }^{ 4 }-2{ r }^{ 3 }-{ r }^{ 2 }+2r+1=0....(iv)$
dividing equation (iv) by ${r}^{2}$ $\Rightarrow { \left( r-\cfrac { 1 }{ r }  \right)  }^{ 2 }-2{ \left( r-\cfrac { 1 }{ r }  \right)  }+1=0\Rightarrow { \left( r-\cfrac { 1 }{ r } -1 \right)  }^{ 2 }=0\Rightarrow \cfrac { r-1 }{ r } =1....(v)\quad $
$\left( \gamma -\alpha  \right) \beta =\left( \alpha { r }^{ 2 }-\alpha  \right) \times \alpha r={ \alpha  }^{ 2 }\left( { \alpha  }^{ 2 }-1 \right) r={ \alpha  }^{ 2 }(r-1)={ \alpha  }^{ 2 }{ r }^{ 2 }$
(using $(v)=\alpha \times \alpha { t }^{ 2 }\quad $

Multiple choice

What is the quadratic formula?

  1. A formula for solving quadratic equations

  2. A formula for solving cubic equations

  3. A formula for solving quartic equations
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The quadratic formula is a formula for solving quadratic equations. It is named after al-Khwarizmi, who derived it in the 9th century.

Multiple choice

Bhaskara II's formula for solving quadratic equations is given by: $$ax^2 + bx + c = 0$$. What is the value of x in this formula?

  1. $$x = (-b ± √(b^2 - 4ac)) / 2a$$
  2. $$x = (-b ± √(b^2 + 4ac)) / 2a$$
  3. $$x = (-b ± √(b^2 - 2ac)) / 2a$$
  4. $$x = (-b ± √(b^2 + 2ac)) / 2a$$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Bhaskara II's formula for solving quadratic equations is given by $$x = (-b ± √(b^2 - 4ac)) / 2a$$. This formula is still used today to solve quadratic equations.

Multiple choice

What was Brahmagupta's formula for solving a quadratic equation?

  1. $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
  2. $x = \frac{-b \pm \sqrt{b^2 + 4ac}}{2a}$
  3. $x = \frac{-b \pm \sqrt{b^2 - 2ac}}{2a}$
  4. $x = \frac{-b \pm \sqrt{b^2 + 2ac}}{2a}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Brahmagupta's formula for solving a quadratic equation is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.