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 maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

Let $\alpha$ and $\beta$ be the roots of the equation ${ x }^{ 2 }+x+1=0$. The equation whose roots are ${ \alpha  }^{ 19 },{ \beta  }^{ 7 }$ is

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

$ { x }^{ 2 }+x+1=0$


$\Rightarrow \left( x-\omega  \right) \left( x-{ \omega  }^{ 2 } \right) =0$

$\Rightarrow x=\omega ,{ \omega  }^{ 2 }$

$\therefore \alpha =\omega ,\beta ={ \omega  }^{ 2 }$   $(\because \omega ,{ \omega  }^{ 2 }$ are cube roots of unity $)$

Hence, ${ \alpha  }^{ 3 }=\omega ^3 =1$
             ${ \beta  }^{ 3 }=[{\omega ^3}]^2 = 1$
             $\alpha \beta =\omega^3=1$

$\therefore { \alpha  }^{ 19 }={ \left( { \alpha  }^{ 3 } \right)  }^{ 6 }\alpha ={ 1 }^{ 6 }\alpha =\alpha =\omega $ and ${ \beta  }^{ 7 }={ \beta  }^{ 6 }.\beta ={ 1 }^{ 2 }.\beta =\beta ={ \omega  }^{ 2 }$

$\\ \Rightarrow { \alpha  }^{ 19 }+{ \beta  }^{ 7 }=\omega +{ \omega  }^{ 2 }=-1$ 
$\Rightarrow { \alpha  }^{ 19 }{ \beta  }^{ 7 }=\omega .{ \omega  }^{ 2 }={ \omega  }^{ 3 }=1$

Hence equation whose roots are ${ \alpha  }^{ 19 },{ \beta  }^{ 7 }$ is

${ x }^{ 2 }-\left( { \alpha  }^{ 19 }+{ \beta  }^{ 7 } \right) x+{ \alpha  }^{ 19 }{ \beta  }^{ 7 }=0$

$\Rightarrow { x }^{ 2 }+x+1$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

Which of the following quadratic equation has the sum of their roots $4$ and the sum of the cubes of their roots as $28$? 

  1. $x^2 - 4x + 3 = 0$
  2. $x^2 - 4x - 5 = 0$
  3. $x^2 - 3x + 4 = 0$
  4. $x^2 + 4x + 3 = 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $\alpha$ and $\beta$ be the roots of the equation.
Hence
$\alpha+\beta=4$
and $\alpha^{3}+\beta^{3}=28$
Now $\alpha^{3}+\beta^{3}$ can be written as

$=(\alpha+\beta)^{3}-3\alpha\beta(\alpha+\beta)$
Hence
$28=64-12\alpha\beta$
$12\alpha\beta=36$
$\alpha\beta=3$
Therefore,
$x^{2}-(\alpha+\beta)x+\alpha\beta=0$
$x^{2}-(4x)+3=0$
Hence, option $A$ is correct.

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $\alpha$ and $\beta$ are the roots of the equation $4x^{2}\, -\, 5x\, +\, 2\, =\, 0$, find the equation whose roots are
$\displaystyle \frac{\alpha^{2}}{\beta}$ and $\displaystyle \frac{\beta^{2}}{\alpha}.$

  1. $2x^{2}\, -\, 5x\, +\, 16\, =\, 0$
  2. $32x^{2}\, +\, 5x\, +\, 16\, =\, 0$
  3. $2x^{2}\, +\, 5x\, +\, 16\, =\, 0$
  4. $32x^{2}\, -\, 5x\, +\, 16\, =\, 0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The given equation is: $4x^2 - 5x + 2 = 0 $


Sum of the roots = $\dfrac{5}{4}$

Product of the roots  = $\dfrac{2}{4} = \dfrac{1}{2}$

If the roots are $\dfrac{\alpha^2}{\beta}, \dfrac{\beta^2}{\alpha}$

Sum of roots = $\dfrac{\alpha^2}{\beta} + \dfrac{\beta^2}{\alpha}$ 

= $\dfrac {\alpha^3 + \beta^3}{\alpha\beta}$

= $\dfrac{{(\alpha + \beta)}^3 - 3\alpha\beta(\alpha+\beta)}{\alpha\beta}$

= $\dfrac{\left (\dfrac{5}{4}\right )^3 - 3 \left (\dfrac{5}{4}\right )\left (\dfrac{1}{2} \right )}{\dfrac{1}{2}}$

= $\dfrac{125 - 120}{32}$

= $\dfrac{5}{32}$

Product of roots = $ \left (\dfrac{\alpha^2}{\beta} \right ) \left ( \dfrac{\beta^2}{\alpha} \right )$

= $\alpha\beta$

= $\dfrac{1}{2}$

Hence.the equation in the standard form, $x^2 - Sx + P  = 0$ can be written as:

=$x^2 - \dfrac{5}{32}x + \dfrac{1}{2} = 0$

= $32x^2 - 5x + 16 = 0$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $\alpha$ and $\beta$ are the roots of the equation $4x^{2}\, -\, 5x\, +\, 2\, =\, 0$, find the equation whose roots are
$\alpha\, +\, 3\beta$ and $3\alpha\, +\, \beta$.

  1. $16x^{2}\, +\, 80x\, +\, 107\, =\, 0$
  2. $16x^{2}\, -\, 80x\, +\, 107\, =\, 0$
  3. $16x^{2}\, -\, 80x\, -\, 107\, =\, 0$
  4. $16x^{2}\, +\, 80x\, -\, 107\, =\, 0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$4x^2 - 5x + 2 = 0 $


If $\alpha$ and $\beta$ are the roots of this equation, 

then , sum of roots: $\alpha + \beta$ = $\displaystyle \frac{5}{4}$

Product of roots: $\alpha. \beta = \displaystyle \frac{2}{4}$

The equation which has roots as : $\alpha + 3\beta$ and $\beta + 3\alpha$

Sum of roots: $4\alpha + 4\beta$ = $4 \left (\dfrac{5}{4} \right ) = 5$

Product of roots: $(\alpha + 3\beta)(3\alpha + \beta) $

$= 3(\alpha^2 + \beta^2) + 10\alpha\beta$

$= 3(\alpha + \beta)^2 - 6\alpha\beta + 10\alpha\beta$

$= 3 \left (\dfrac{5}{4} \right )^2 + 4\frac{2}{4}$

$= \dfrac{107}{16}$

Thus new equation is :$x^2 -Sx + P = 0$


$\therefore x^2 - 5x + \dfrac{107}{16} = 0$

$\therefore 16x^2 - 80x + 107 = 0$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If one root of the quadratic equation $ax^{2}\, +\, bx\, +\, c\, =\, 0$ is the square of the other, then $b^{3}\, +\, a^{2}c\, +\, ac^{2}\, =\, 3abc$
Say yes or no.

  1. Yes

  2. No

  3. Ambiguous

  4. Data insufficient
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let one root of $ax^2 + bx + c =0$ be $\alpha$ and other be $\alpha^2$
then, $\alpha + \alpha^2 = \frac{-b}{a}$
$\alpha^3 = \frac{c}{a}$
or $\alpha = (\frac{c}{a})^{({\frac{1}{3}})}$
Now put the value of $\alpha$ in $\alpha + \alpha^2 = \frac{-b}{a}$
$\frac{c}{a}^{\frac{1}{3}} + \frac{c}{a}^{\frac{2}{3}} = \frac{-b}{a}$
Cubing both sides:

$(\frac{c}{a})^{({\frac{3}{3}})} + (\frac{c}{a})^{({\frac{6}{3}})} + 3 {(\frac{c}{a})^{({\frac{1}{3}})}}\times{(\frac{c}{a})^{({\frac{2}{3}})}}((\frac{c}{a})^{({\frac{1}{3}})} + (\frac{c}{a})^{({\frac{2}{3}})}) = (\frac{-b}{a})^3$

$\frac{c}{a} + \frac{c^2}{a^2} + 3\frac{c}{a}(\frac{-b}{a}) = \frac{-b^3}{a^3}$

$a^2c + ac^2 - 3abc = - b^3 $
$b^3 + a^2c + ac^2 = 3abc$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If the roots of the equation $2x^2 - 3x + 5 = 0$ are reciprocals of the roots of the equation $ax^2 + bx + 2 = 0$, then

  1. $a = 2, b = 3$
  2. $a = 2, b = -3$
  3. $a = 5, b = -3$
  4. $a = 5, b = 3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $\alpha ,\beta $ are roots of $2{ x }^{ 2 }-3x+5=0$ 
Then to get equation whose roots are $\displaystyle \dfrac { 1 }{ \alpha  } ,\dfrac { 1 }{ \beta  } $ 
Replace $\displaystyle x\rightarrow \frac { 1 }{ x } \ $
We get $5{ x }^{ 2 }-3x+2=0$.

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If each root of the equation ${x}^{2}+11{x}+13=0$ is diminished by $4$, then the resulting equation is

  1. ${x}^{2}+3{x}-15=0$
  2. ${x}^{2}+3{x}+73=0$
  3. ${x}^{2}+19{x}+73=0$
  4. ${x}^{2}-3{x}-4=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$ The\quad roots\quad of\quad the\quad equation\quad { x }^{ 2 }+11x+13=0\quad is\quad diminished\quad by\quad 4\quad i.e.\ the\quad variable\quad becomes\quad (x+2).\ \therefore \quad
The\quad new\quad equation\quad is\quad \ (x+4)^{ 2 }+11(x+4)+13=0\ \Rightarrow { x }^{ 2 }+8x+16+11x+44+13=0\ \Rightarrow { x }^{ 2 }+19x+73=0\ Ans-\quad Option\quad C .$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $\displaystyle \alpha, \beta $ are the roots of $\displaystyle x^{2}+3x+3=0$  then find the quadratic equation whose roots are $\displaystyle (\alpha +\beta )$ and $\displaystyle \alpha \beta $

  1. $\displaystyle x^{2}=1$
  2. $\displaystyle x^{2}=4$
  3. $\displaystyle x^{2}=9$

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

For the equation, sum of roots $ = \alpha  + \beta = -\dfrac {3}{1} = -3 $
Product of roots $ = \alpha  \times \beta = \dfrac {3}{1} = 3 $

So, the eqn with roots $ = \alpha  + \beta$ and $ \alpha  \times \beta $ is $ (x - (-3))(x-3) = 0 $
$ => x^{2} -9 = 0 $

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $a, b, g$  are the roots of the equation $(x - 2$ ) $\displaystyle \left ( x^{2}+6x-11 \right )=0$ therefore $(a + b + g)$  equals

  1. $-4$
  2. $\dfrac{23}{6}$
  3. $13$
  4. $-8$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, $(x-2)(x^{2}+6x-11)=0$
$x^{3}+6x^{2}-11x-2x^{2}-12x-22=0$
$x^{3}+4x^{2}-23x-22=0$
Then $a=1  ,b=4  g=-22$
Sum of roots $(a+b+g) =\displaystyle \frac{-b}{a}=\frac{-4}{1}=-4$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

The roots of equation $\displaystyle x^{2}+px+q=0$ are $1 $ and $2$ . The roots of the equation $\displaystyle qx^{2}-px+1=0$ must be

  1. $-1,$ $\displaystyle -\frac{1}{2}$
  2. $\displaystyle \frac{1}{2},1$
  3. $\displaystyle -\frac{1}{2},1$
  4. $\displaystyle -1,\frac{1}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation can be written as $x^{ 2 }+px+q=0$

The roots are $1$ and $2$
Sum of roots $= 3= -p$
Product of roots $= 2= q$
The second equation is $2x^{ 2 }+3x+1=0$
$2x^{ 2 }+2x+x+1=0$
$ \Longrightarrow 2x(x+1)+1(x+2)=0$
$\Longrightarrow (x+1)(2x+1)=0$
$ \Longrightarrow x=-1 $ or $-1/2$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

The equation whose roots are twice the roots of $x^2 -3x +3=0$ is

  1. $x^2-6x+12=0$
  2. $x^2-3x+6=0$
  3. $2x^2-3x+3=0$
  4. $4x^2-6x+3=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$x^2 -3x = 3 = 0$ ........ (1)
Here, $a +\beta =3$ and $a\beta =3$. Therefore,
$2(a +\beta ) =6$
$2\times 2a\beta = 4a\beta =4 \times 3= 12$
The equation whose roots are double of (1),
$x^2 -$(sum of the roots)x + product of the roots $=0$
will be $x^2 -6x + 12 =0$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

The equation whose roots are the squares of the roots of equation $x^2 -x +1= 0$ is

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

The given equation is $x^2 -x + 1 = 0$ ....... (1)
Here, $a +\beta = 1$ and $a\beta = 1$. Therefore,
$a^2+\beta^2 = (a + \beta)^2 -2a\beta = 1-2= -1$
and $a^2\beta^2 = (a\beta)^2 = 1^2= 0$
Therefore, the equation whose roots are square of 1 is $x^2$ -(sum of the roots)x +product $=0$
or $x^2-(-1)x+ 1 =0$
or $x^2+x+ 1 =0$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $m$ and $n$ are the roots of the equation $(x + p)(x + q) - k = 0$, then the roots of the equation $(x - m)(x - n) + k = 0$ are-

  1. $p$ and $q$
  2. $1/p$ and $1/q$
  3. $-p$ and $-q$
  4. $p + q$ and $p - q$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$(x+p)(x+q)-k=0\ \Longrightarrow { x }^{ 2 }+(p+q)x+pq-k=0$

$m$ and $n$ are the roots of this equation
So, we have
Sum of roots $= -(p+q)=m+n$
Product of the roots $=pq-k= mn$
$\Rightarrow pq=mn+k$
Consider, $(x-m)(x-n)+k=0$ 
$\Rightarrow { x }^{ 2 }-(m+n)x+mn+k=0$
Sum of roots is $ m+n$
But $m+n= (-p)+(-q)$
Product of the roots $=mn+k$
But $mn+k= pq= (-p)(-q)$
Hence, the roots of the new equation are $-p,-q$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $\alpha$ and $\beta$ are the roots of $x^{2} + p = 0$ where p is a prime, which equation has the roots $\dfrac {1}{\alpha}$ and $\dfrac {1}{\beta}$?

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

${ x }^{ 2 }+p=0$

roots are $\alpha & \beta $
sum = $\alpha +\beta =0$
product=$\alpha \beta =p$
New roots are $\cfrac { 1 }{ \alpha  } & \cfrac { 1 }{ \beta  } $
sum = $\cfrac { 1 }{ \alpha  } +\cfrac { 1 }{ \beta  } =\cfrac { \alpha +\beta  }{ \alpha \beta  } =0\ $
product = $\cfrac { 1 }{ \alpha \beta  } =\cfrac { 1 }{ p } $
equation 
${ x }^{ 2 }$-(sum of roots)x+product of roots = 0
${ x }^{ 2 }-0+\cfrac { 1 }{ p } =0\ { px }^{ 2 }+1=0$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

The equation formed by multiplying each root of $ax^2  + bx + c = 0$ by 2 is $ x^2 + 36x + 24 = 0$.Which one of the following is correct ?

  1. $ bc = a^2 $
  2. $ bc = 36 a^2 $
  3. $ bc = 72 a^2 $
  4. $ bc = 108 a^2 $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

let $p,q$ be roots of equation $ax^2+bx+c=0$


So $p+q=\left(-\dfrac{b}{a}\right)$ and $pq=c/a$


$\Rightarrow b=-a(p+q),c=apq$

New equation is $x^2+36x+24=0$ and roots are $2p,2q$

So $2p+2q=-36$

$\Rightarrow p+q=-18$

$2p\times 2q=24$

$\Rightarrow pq=6$

Then, value of $bc$ is $[-a(p+q)][apq]=-a(-18)a6=108a^2$