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 algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

If the sum of two of the roots of $x^4-2x^3-3x^2+10x-10=0$ is zero then the roots are

  1. $\pm \sqrt{5},1\pm i$
  2. $\pm \sqrt{5},1-i$
  3. $\large{\frac{1}{2}},-\large{\frac{1}{5}},\pm 1$
  4. $\sqrt{2},\sqrt{5},\pm 2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

let roots are $\pm a,b,c\ b+c=2\ -{ a }^{ 2 }bc=-10\ { a }^{ 2 }bc=10\ { -a }^{ 2 }+ab+ac+bc-ab-ac=-3\ bc-{ a }^{ 2 }=-3\ { a }^{ 2 }-bc=3\ $

let $bc=t$
from $(2) t=\frac { 10 }{ { a }^{ 2 } } \ (3)\quad \quad { a }^{ 2 }-t=3\ { a }^{ 2 }-\frac { 10 }{ { a }^{ 2 } } =3\ { a }^{ 4 }-{ 3a }^{ 2 }-10=0\ { a }^{ 4 }-{ 5a }^{ 2 }+{ 2a }^{ 2 }-10=0\ { a }^{ 2 }\left( { a }^{ 2 }-5 \right) +2\left( { a }^{ 2 }-5 \right) =0\ \left( { a }^{ 2 }-5 \right) \left( { a }^{ 2 }+2 \right) =0\ a=\pm \sqrt { 5 } \ bc=2\ c=\cfrac { 2 }{ b } \ b+\cfrac { 2 }{ b } =2\ { b }^{ 2 }-2b+2=0$
$\quad \quad b = 1 \pm i$
$ \quad \quad c = \cfrac{2}{b} = \cfrac{2}{1\pm i} = 1 \mp i$
$ \therefore $ roots are $ 1\pm i, \pm\sqrt5$

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

If two roots of the equations $x ^ { 3 } - p x ^ { 2 } + q x - r = 0$ are equal in magnitude but opposite in sign, for

  1. pr = q

  2. qr = p

  3. pq = r
  4. $p ^ { 2 } q ^ { 2 } = r$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation 
let those are m, -m
now sum of three roots = p
hence third root will be p
now 
m*(-m) + m*p + (-m)*p = q
hence  –m2 = q
now m*( –m) * p = r
 –m2 p  = r
put value of  –m2 = q
hence  pq = r

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

If the equation ${x}^{4}-4{x}^{3}+a{x}^{2}+bx+1=0$ has four positive roots, then the value of $(a+b)$ is:

  1. $-4$
  2. $2$
  3. $6$

  4. cannot be determined
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $x^4 - 4x^3 + ax^2 + bx + 1 = 0$

let the root of equation be $\alpha, \beta, \gamma, \sigma$
$\alpha + \beta + \gamma + \sigma = 4$ ...(i)
$\alpha \beta \gamma \sigma = 1$ ... (ii)
$\dfrac{1}{4} (\alpha + \beta + \gamma + \sigma) = 1$
$\Rightarrow \dfrac{1}{4} (\alpha + \beta + \gamma + \sigma) = (\alpha \beta \gamma \sigma) \dfrac{1}{4}$
$\therefore A. M. = a. m.$
$\therefore \alpha = \beta = \gamma = \sigma$
$4 \alpha = 4$
$\therefore \alpha = 1$
$1 - 4 + a + b + 1 = 0$
$\therefore a + b = 2$

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

The value of $'a'$ for which the equation ${ x }^{ 3 }+ax+1=0$ and ${ x }^{ 4 }+a{ x }^{ 2 }+1=0$, have a common root is

  1. $a=2$
  2. $a=-2$
  3. $a=0$

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

Consider the following equation
$x^{4}+ax^{2}+1=0$
$x^{3}+ax+1=0$
Subtracting equation (ii) from (i), we get 
$x^{4}-x^{3}+a(x^{2}-x)=0$
$x^{3}(x-1)+ax(x-1)=0$
$(x-1)(x^{3}+ax)=0$
$x(x-1)(x^{2}+a)=0$
Hence, we get $x=0$ $x=1$ and $x^{2}=-a$
Now out of the above two, $x=0$ is not a root of the following two equations.
We do not know the nature of '$a$'. 

Hence, we cannot determine that $x^{2}=-a$ will have real or imaginary roots.
Hence, we get $x=1$ as a common root for the above two equations.
Now for both the equations to have $x=1$ as a common root, 
$f(1)=0$
$1+a+1=0$
$a=-2$
Similarly substituting in the second equation, we get $a=-2$.
Hence, the required value of $a$ is $-2$.

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

lf the difference of the roots of the equation $x^{2}-bx+c=0$ is equal to the differecne of the roots of the equation ${x}^{2}-{c}x+b=0$ and $b\neq c$, then $b+c=$

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

Let $(\alpha, \beta )$ and $ (\gamma, \delta )$ be the roots of the first equation and second equation respectively. 

Then, for the first equation
$ \alpha +\beta =b$ and $ \alpha \beta =c$
Now $ (\alpha +\beta )^{ 2 }={ b }^{ 2 }$
$ \Rightarrow (\alpha -\beta )^{ 2 }+4\alpha \beta ={ b }^{ 2 }$
$ \Rightarrow (\alpha -\beta )^{ 2 }={ b }^{ 2 }-4\alpha \beta $
$ \Rightarrow |\alpha -\beta |=\sqrt { { b }^{ 2 }-4c } $
Similarly for the second equation
$ |\gamma -\delta |=\sqrt { { c }^{ 2 }-4b } $
As per the given condition,
$ \sqrt { { b }^{ 2 }-4c } =\sqrt { { c }^{ 2 }-4b } $
$\Rightarrow { b }^{ 2 }-4c={ c }^{ 2 }-4b$
$\Rightarrow { b }^{ 2 }-{ c }^{ 2 }=-4(b-c)$
$ \Rightarrow (b+c)(b-c)=-4(b-c)$
Therefore, option D is correct.

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

Let $\displaystyle a _{1}, a _{2},a _{3},a _{4},a _{5} \, \varepsilon \, R$ denote a rearrangement of equation $\displaystyle p _{1}x^{5}+p _{2}x^{3}+p _{3}x^{2}+p _{4}x+p _{5}=0$ then, equation  $\displaystyle a _{1}x^{4}+a _{2}x^{3}+a _{3}x^{2}+a _{4}x +a _{5}=0$ has 

  1. at least two real roots

  2. all four real roots

  3. only imaginary roots

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$a _{1}x^{4}+a _{2}x^{3}+a _{3}x^{2}+a _{4}x+a _{5}=0$
for $ x=1 $
$a _{1}+a _{2}+a _{3}+a _{4}+a _{5}=0$
for the given set of equation,
sum of  $p _{1}+p _{2}+p _{3}+p _{4}+p _{5}=0$ 
So, $a _{1}+a _{2}+a _{3}+a _{4}+a _{5}\epsilon R $  for any set of arrangement
Hence at least two real roots.
Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

If the sum of two roots of the equation $x^{4}+px^{3}+qx^{2}+rx+8=0$ is equal to the sum of the other two, then $p^{3}+8r=$

  1. $p^2 - 4pq$
  2. $2pq$
  3. $p^2 - pq$
  4. $4pq$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the roots of the equation be $a,b,c,d$. 
As per the question,
$a+b = c+d$ 
From the theory of polynomials, 
$a+b+c+d = -p$
$ \Rightarrow a+b=c+d= \displaystyle \frac{-p}{2} $

Also,
$ab+ac+ad+bd+bc+cd  = q $
$ \Rightarrow (a+b)(c+d) +ab+cd  =q $
$ \Rightarrow ab+cd = q - \displaystyle \frac{p^2}{4} $

Also, 
$abc+abd+bcd+adc = -r $
$ \Rightarrow ab(c+d) +cd(a+b) = -r $
$ \Rightarrow \displaystyle \frac {-p}{2} (ab+cd) = -r $
$ \Rightarrow \displaystyle \frac {-p}{2} ( q - \displaystyle \frac{p^2}{4} ) = -r $
$ \Rightarrow -4pq + p^3 = -8r $
$ \Rightarrow p^3 + 8r = 4pq $

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

lf one root of the equation $ax^{2}+bx+c=0$ is the square of the other, then

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

Given equation $a{ x }^{ 2 }+bx+c$
Given that, one root of the equation is square of another.
So, lets assume $\alpha$ , ${ \alpha  }^{ 2 }$ are roots of the given equation
We know that,
Sum of roots $=$ $\alpha +{ \alpha  }^{ 2 }=\dfrac { -b }{ a }$ 
Product of roots $=$ $ \alpha \times { \alpha  }^{ 2 }=\dfrac { c }{ a }$
$\alpha (1+\alpha )=\dfrac { -b }{ a } \longrightarrow 1  $
${ \alpha  }^{ 3 }=\dfrac { c }{ a } \longrightarrow 2 $
Cubing equation (1) on both sides and substitute the value from equation (2).
${ \alpha  }^{ 3 }{ (1+\alpha ) }^{ 3 }=\dfrac { -{ b }^{ 3 } }{ { a }^{ 3 } } \ { \alpha  }^{ 3 }({ \alpha  }^{ 3 }+1+3{ \alpha  }(1+\alpha ))=\dfrac { -{ b }^{ 3 } }{ { a }^{ 3 } } \ \dfrac { c }{ a } \left (\dfrac { c }{ a } +1+3\left (\dfrac { -b }{ a } \right)\right)=\dfrac { -{ b }^{ 3 } }{ { a }^{ 3 } } \ \dfrac { ({ c }^{ 2 }+ac-3bc) }{ { a }^{ 2 } } =\dfrac { -{ b }^{ 3 } }{ { a }^{ 3 } } \ a({ c }^{ 2 }+ac-3bc)=-{ b }^{ 3 }\ { b }^{ 3 }+a{ c }^{ 2 }+{ a }^{ 2 }c=3abc $

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

If $p, q, r, s, t$ are the roots of the equation $x^5-1 = 0$, then $p^{ 10 }+q^{ 10 }+{ r }^{ 10 }+{ s }^{ 10 }+t^{ 10 }=$

  1. $0$
  2. $1$
  3. $3$
  4. $5$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$p, q, r, s, t$ are all fifth root of unity
$\Rightarrow p^5=q^5= r^5= s^5= t^5=1$   
$ \Rightarrow p^{10}=q^{10}= r^{10}= s^{10}= t^{10}=1$
Hence, the required sum is $5$

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

The maximum number of real root of the equation $\displaystyle x^{2n} - 1 = 0$ is

  1. $\displaystyle 2$
  2. $\displaystyle 3$
  3. $\displaystyle n$
  4. $\displaystyle 2n$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$x^{2n}=1$
Now if  $n$ is odd we have
$x^{n}=\pm1 $
$x^{n}=1$ and $x^{n}=-1$
$x^{n}=-1$
$x=-1$
Now if $n$ is odd
$x^{n}-1=0$
$(x-1)(1+x+x^{2}+..x^{n-1})=0$
Hence $x=1$ and  the equation $1+x+x^{2}+..x^{n-1}=0$ gives $nth$  roots of unity.
Hence at most $2$ real roots.
Similarly if $n$  is even.
Then
$x^{n}=\pm1 $
$x^{n}=1$ and $x^{n}=-1$
Now 
$x^{n}=-1$ will given imaginary roots.
$x^{n}=1$ can be further simplified in to
$x^{\frac{n}{2}}=\pm1 $ and so on.
Hence we will get remaining pairs of imaginary roots ans two real roots $1$ and  $-1$  at the end.
Hence at-most $2$ real roots.

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