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 applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

The number of roots of the equation  $\displaystyle x-\frac{2}{(x-1)}=1-\frac{2}{(x-1)}$ is 

  1. 0

  2. 1

  3. 2

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

Given, $ x - \cfrac {2}{(x-1)} = 1 - \cfrac {2}{(x-1)} $

Cancelling out $ - \cfrac {2}{(x-1)} $ from LHS and RHS we get, $ x = 1 $
But when $ x = 1 $, denominator of fraction $ - \cfrac {2}{(x-1)} $ is $ 0 $, which is not defined.
Hence, there is no root of this equation.

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

If $(x - a) (x - 5) + 2 = 0$ has only integral roots where $\displaystyle a \, \varepsilon \, I,$ then the value of $a$ can be 

  1. $8$
  2. $7$
  3. $6$
  4. $5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $\alpha, \beta $ be the roots of the given equation


$\alpha +\beta =5+a$

$\alpha ×\beta=5a+2$

Eliminating "a" from the equation, we get ;

$\Rightarrow 5(\alpha +\beta) - \alpha \beta =23$

Here we have to choose the pairs $\alpha, \beta$ so that it satisfies above equation 

By trial and error method, we find $\alpha =7,\beta=6$ or vice versa 

Substituting $\alpha, \beta$ in above equation, we get a=8

$\therefore a=8$

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

If the roots of the equation $\displaystyle px^{2}+qx+r=0$ are in the ratio $\displaystyle \varphi \ : \ m,$ then 

  1. $\displaystyle (\varphi +m)^{2}qp=\varphi mr^{2}$
  2. $\displaystyle (\varphi +m)^{2}pr=\varphi mq$
  3. $\displaystyle (\varphi +m)^{2}pr=\varphi mq^{2}$

  4. None of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let the roots be $\varphi r$ and $mr$
Sum of roots $= \varphi r+mr=-\dfrac{q}{p}$
$\therefore  (\varphi+m)r=-\dfrac{q}{p}$
$\therefore (\varphi+m)^2r^2=\dfrac{q^2}{p^2}$   ...(1)

Product of roots $= (\varphi r)(mr)=\dfrac{r}{p}$
$\therefore  \varphi mr^2=\dfrac{r}{p}$    ...(2)

Dividing equation (1) by (2), we get

$\dfrac{(\varphi+m)^2r^2}{\varphi mr^2}=\dfrac{\frac{q^2}{p^2}}{\frac{r}{p}}$

$\therefore \dfrac{(\varphi+m)^2}{\varphi m}=\dfrac{q^2}{pr}$

$\therefore (\varphi+m)^2pr=\varphi mq^2$
Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

For the equation $|x|^{2}+|x|-6=0$, the roots are

  1. one and only one real number.

  2. real with sum one.

  3. real with sum zero.

  4. real with product zero.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For x>0 equation is $x^2+x-6=0$
$(x-2)(x+3)=0$
$x=2$
$x$ cant be equal to -3 as for this equation $x>0$
Now when $x <0$ equation becomes $x^2-x-6$
$(x+2)(x-3)=0$
Hence $x=-2$
So the roots are $2 and -2$
Thus sum of roots is zero and roots are real
So Option C

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

If $a, b, c$ are in A.P., then the roots of the equation $ax^{2}+2bx+c=0$ are

  1. real and distinct

  2. real and equal

  3. real

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

If $a,b,c$ are in A.P, then

$2b=a+c$.
Hence the give equation transforms into
$ax^{2}+(a+c)x+c=0$
Hence
$D$
$=B^{2}-4AC$

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

$=a^{2}+c^{2}+2ac-4ac$

$=a^{2}+c^{2}-2ac$

$=(a-c)^{2}$
Now 
$(a-c)^{2}\geq 0$
Hence 
$D\geq 0$.
Or 
$B^{2}-4AC\geq 0$.
Since the discriminant is greater than 0, hence the roots real.

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

If $a$ and $b$ are the roots of the quadratic equation $x^2-4x+3=0$, then $(1+a+a^2+a^3...)(1+b+b^2+b^3+....)$ equal to

  1. $\infty$
  2. $\dfrac{1}{4}$
  3. $\dfrac{-1}{6}$

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

${ x }^{ 2 }-4x+3=0\ (x-1)(x-3)=0\ x=1\quad or\quad 3\ \therefore a=3,\quad b=1\ a+b=4,\quad ab=3\ 1+a+{ a }^{ 2 }+{ a }^{ 3 }+........+\infty =\cfrac { 1 }{ 1-a } (sum\quad of\quad infinite\quad G.P.)\ \therefore (1+a+{ a }^{ 2 }+{ a }^{ 3 }+........+\infty )(1+b+{ b }^{ 2 }+{ b }^{ 3 }+.......+\infty )\ =(\cfrac { 1 }{ 1-a } )(\cfrac { 1 }{ 1-b } )\ =\cfrac { 1 }{ 1-(a+b)+ab } \ =\cfrac { 1 }{ 1-4+3 } =\cfrac { 1 }{ 0 } =\infty $

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

If $a,b,c,d$ are four consecutive terms of an increasing A.P., then the roots of the equation
$(x-a)(x-c)+2(x-b)(x-d)=0$ are

  1. $\text{real and distinct}$
  2. $\text {non-real complex}$
  3. $\text {real and equal}$
  4. $\text {integers}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$a,b,c,d$ are $4$ consecutive terms of A

Let $a=m-3n,b=m-n.c=m+n,d=m+3n$

$'2n'$ is common difference

$(x-a)(x-c)+2(x-b)(x-d)=0$

$3{ x }^{ 2 }-\left( a+c+2b+2d \right) x+\left( ac+2bd \right) =0$

$=6m+2n$

$ac+2bd={ m }^{ 2 }-2mn-3{ n }^{ 2 }+2{ m }^{ 2 }+4mn-6{ n }^{ 2 }$

$=3{ m }^{ 2 }+2mn-9{ n }^{ 2 }$

$3{ x }^{ 2 }-\left( 6m+2n \right) x+\left( 3{ m }^{ 2 }+2mn-9{ n }^{ 2 } \right) =0$

$\triangle ={ \left( 6m+2n \right)  }^{ 2 }-4\left( 3 \right) \left( 3{ m }^{ 2 }+2mn-9{ n }^{ 2 } \right) $

$=4\left( 9{ m }^{ 2 }+6mn+{ n }^{ 2 }-9{ m }^{ 2 }-6mn+36{ n }^{ 2 } \right) $

$=4\left( 37 \right) { n }^{ 2 }$

$\triangle >0$

$\therefore $ Roots of $(x-a)(x-c)+2(x-b)(x-d)$ are real and distinct.
Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

If roots of equation $ x^2 - (2n+ 18) x - n-1 = 0 ( n \epsilon Z ) $ are rational, then number of possible value of $n $ is :

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

  4. Infinite
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For roots to be rational, the discriminant D = (2n+18)^2 - 4(1)(-n-1) must be a perfect square. D = 4n^2 + 72n + 324 + 4n + 4 = 4n^2 + 76n + 328. Setting 4n^2 + 76n + 328 = k^2, we find integer solutions for n.

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

 Choose the correct answer from the alternatives given.
If $\alpha \, and \, \beta$ are the roots of the equation $x^2$ - 7x + 12 = 0, then $\alpha^2 \, + \, \beta^2$ equals.

  1. 19

  2. 25

  3. 14

  4. 24
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
: Let $\alpha \, and \, \beta$ are the roots of the equation
$ax^2 + bx + c = 0$
We know that,
$\displaystyle \alpha \, + \, \beta \, = \, \frac{-b}{a} \, = \, \frac{- (-7)}{1} \, = \, 7$
$\displaystyle \alpha^2 \, + \, \beta^2 \, = \, (\alpha \, + \, \beta)^2 \, - \, 2\alpha \beta$
$\displaystyle \alpha^2 \, + \, \beta^2 \, = \, (7)^2 \, - \, 2 \, \times \, 12 \, = \, 49 \, - \, 24 \, = \, 25$
Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

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

$Clearly\quad x=1\quad is\quad a\quad solution$
$\therefore \quad Product\quad of\quad the\quad roots=\frac { a-b }{ b-c } $$\therefore \quad \left( 1 \right) \left( 1 \right) =\frac { a-b }{ b-c } $
$\Rightarrow b-c=a-b$
$\Rightarrow 2b=a+c\Rightarrow a,b,c\quad are\quad in\quad A.P.$






Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

If the harmonic mean of the roots of$\sqrt { 2 } { x }^{ 2 }-bx+\left( 8-2\sqrt { 5 }  \right) =0$ is 4, the the value of b=

  1. 2

  2. 3
  3. $4-\sqrt { 5 } $
  4. $4+\sqrt { 5 } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$Let\quad \alpha ,\beta \quad be\quad the\quad roots$ 
$\Rightarrow \frac { 2\alpha \beta  }{ \alpha +\beta  } =4\quad \quad \Rightarrow \frac { \frac { 2\left( 8-2\sqrt { 5 }  \right)  }{ \sqrt { 2 }  }  }{ \frac { b }{ \sqrt { 2 }  }  } =4$
$\Rightarrow \frac { 2\left( 8-2\sqrt { 5 }  \right)  }{ 4 } =b$
$\therefore b=4-\sqrt { 5 } $




Multiple choice the nth roots of unity complex numbers maths

If $1,\alpha, \alpha^2,.....,\alpha^{n - 1}$ be the $n^{th}$ roots of unity, then $(1-\alpha)(1-\alpha^2).....(1-\alpha^{n-1}) $

  1. $3$
  2. $0$
  3. $n$
  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Basically $1,a^1,a^2......a^{n-1}$ all these are the roots of this equation $x^3 – 1 =0$
So we can write
$x^3 -1= (x-1)(x-a _1)(x-a _2).....(x-a _{n-1})$

$\dfrac{x^3 -1}{(x-1)}= (x-a^1)(x-a^2).....(x-a^{n-1})$

$x^2 + x +1= (x-a^1)(x-a^2).....(x-a^{n-1})$

Put $x=1$ on both the sides now

Ans $=3$
Multiple choice the nth roots of unity complex numbers maths

If $\alpha _1, \alpha _2, \alpha _3, \alpha _4$ be the roots of $x^5 - 1 = 0$ then find $\displaystyle \frac{\omega - \alpha _1}{\omega^2 - \alpha _1} \cdot \frac{\omega - \alpha _2}{\omega^2 - \alpha _2} \cdot \frac{\omega - \alpha _3}{\omega^2 - \alpha _3} \cdot \frac{\omega - \alpha _4}{\omega^2 - \alpha _4} $

  1. $\omega^2$
  2. $1$
  3. $\omega$
  4. $(\omega-\alpha _1)(\omega-\alpha _2)(\omega-\alpha _3)(\omega-\alpha _4)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$x^5 - 1 = 0$ has roots $1, \alpha _1, \alpha _2, \alpha _3, \alpha _4$
$\therefore (x^5 - 1) = (x- 1) (x - \alpha _1) (x - \alpha _2) (x - \alpha _3) (x- \alpha _4)$
$\Rightarrow \displaystyle \frac{x^5 -1}{x - 1} = (x - \alpha _1) (x- \alpha _2) (x - \alpha _3) (x - \alpha _4)$           ........   (1)
Putting $x = \omega$ (1) we have
$\displaystyle \frac{\omega^5 - 1}{\omega - 1} = (\omega - \alpha _1) (\omega - \alpha _2) (\omega - \alpha _3) (\omega - \alpha _4)$
$\displaystyle \frac{\omega^2 - 1}{\omega - 1} = (\omega - \alpha _1) (\omega - \alpha _2) (\omega - \alpha _3) (\omega - \alpha _4)$         ....... (2)
and putting $x = \omega^2$ in (1) we have
$\displaystyle \frac{\omega^{10} - 1}{\omega^2 - 1} = (\omega^2- \alpha _1) (\omega^2 - \alpha _2) (\omega^2 - \alpha _3) (\omega^2 - \alpha _4)$
$\Rightarrow \displaystyle \frac{\omega - 1}{\omega^2 - 1} = (\omega^2- \alpha _1) (\omega^2 - \alpha _2) (\omega^2 - \alpha _3) (\omega^2 - \alpha _4)$           ....... (3)
Dividing (2) by (3)
then $\displaystyle \frac{\omega - \alpha _1}{\omega^2 - \alpha _1} \cdot \frac{\omega - \alpha _2}{\omega^2 - \alpha _2} \cdot \frac{\omega - \alpha _3}{\omega^2 - \alpha _3} \cdot \frac{\omega - \alpha _4}{\omega^2 - \alpha _4} \cdot = \frac{(\omega^2 - 1)^2}{(\omega - 1)^2}$
                                                                                  $= \displaystyle \frac{\omega^4 + 1 - 2 \omega^2}{\omega^2 + 1 - 2 \omega}$
                                                                                   $= \displaystyle \frac{\omega + 1 - 2 \omega^2}{\omega^2 + 1 - 2 \omega}$
                                                                                   $= \displaystyle \frac{- \omega^2 - 2 \omega^2}{- \omega - 2 \omega}$
                                                                                    $= \displaystyle \frac{- 3 \omega^2}{- 3 \omega}$
                                                                                    $= \omega$

Ans: C

Multiple choice the nth roots of unity complex numbers maths

If $a = cos \dfrac{2\pi}{7}+i  sin\dfrac{2\pi}{7}$, then find the quadratic equation whose roots are $a = a + a^2 + a^4$ and $\beta = a^3 + a^5 + a^6$.

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

$a = cos (2\pi/7)+i  sin(2\pi/7)$
$\Longrightarrow a^7 = [cos(2\pi/7)+i  sin(2\pi/7)]^7$
$= cos 2\pi + i sin 2\pi = 1$          (1)
$S = \alpha + \beta = (a + a^2 + a^4) + (a^3 + a^5 + a^6)$
$= a +a^2 +a^3 +a^4 + a^5 +a^6 = \frac{a(1-a^6)}{1-a}$
$= \frac{a-a^7}{1-a} = \frac{a-1}{1-a} = -1$                           (2)
$P= \alpha \beta = (a+a^2+a^4)(a^3+a^5+a^6)$
$= a^4+a^6 +a^7+a^5+a^7+a^8+a^7+a^9+a^{10}$
$= a^4+a^6+1+a^5+1+a+1+a^2+a^3$         [From Eq. (1)] 
$= 3+(a+ a^2+ a^3+ a^4+ a^5 + a^6)$ 
$= 3+S = 3-1=2$              [From Eq. (2)]
Therefore, the required equation is 
$x^2 -Sx + P = 0$
$\Longrightarrow x^2 + x + 2=0$


Ans: D