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

If $\alpha , \beta$ are the roots of the equation $ax^2+bx+c=0$ then the quadratic equation whose roots are $\alpha + \beta , \alpha \beta$ is:

  1. $a^2 x^2 +a(b-c) x+bc=0$
  2. $a^2 x^2 + a(b-c) x-bc=0$
  3. $ax^2 +(b+c) x+bc=0$
  4. $ax^2-(b+c)x-bc=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

From the first equation, we can conclude that
$\alpha+\beta=-\cfrac{b}a$  ...(i)

$\alpha\beta=\cfrac{c}a$      ...(ii)

Therefore, the new equation will be

$x^2-(\alpha+\beta+\alpha\beta)x+(\alpha+\beta)(\alpha\beta)=0$

Substituting the values from (i) and (ii), we get

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

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


Hence, the answer is
$a^2x^2+a(b-c)x-bc=0$

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 $ax^2+bx+c=0$ and if $px^2+qx+r=0$ has roots $\displaystyle \frac{1-\alpha}{\alpha}$ and $\displaystyle \frac{1-\beta}{\beta}$, then $r$ is

  1. $a+2b$
  2. $a+b+c$
  3. $ab+bc+ca$
  4. $abc$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
The equation with roots $\cfrac1{\alpha}$ and $\cfrac1{\beta}$
$=a\left(\cfrac1x\right)^2+b\left(\cfrac1x\right)+c$
$=cx^2+bx+a=0$ ..(1)
Now $\cfrac{1-\alpha}{\alpha}=\cfrac{1}{\alpha}-1$
Similarly $\cfrac{1-\beta}{\beta}=\cfrac{1}{\beta}-1$
Therefore the quadratic equation containing these roots is
$c\left(x+1\right)^2+b\left(x+1\right)+a$
$=cx^2+\left(b+2c\right)x+a+b+c = 0$
By comparing coefficients we get with $px^2+qx+r=0$ we get
$r=a+b+c$
Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $\alpha , \beta$ are the roots of the equation $9x^2+6x+1=0$, then the equation with the roots $\cfrac{1}{\alpha}, \cfrac{1}{\beta}$ is :

  1. $2x^2+3x+18=0$
  2. $x^2+6x-9=0$
  3. $x^2+6x+9=0$
  4. $x^2-6x+9=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$\left(x-\cfrac1{\alpha}\right)\left(x-\cfrac1{\beta}\right)=0$
$x^2-\left(\cfrac{1}{\alpha}+\cfrac{1}{\beta}\right)x+\left(\cfrac{1}{\alpha\beta}\right)=0$
$x^2-\left(\cfrac{\alpha+\beta}{\alpha\beta}\right)x+\left(\cfrac{1}{\alpha\beta}\right)=0$
From the given equation we know
$\alpha+\beta=-\cfrac69$
$\alpha\beta=\cfrac19$
By substituting we get
$x^2-\left(-6\right)x+9=0$
$x^2+6x+9=0$
Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $\alpha$ and $\beta$ are roots of $2{ x }^{ 2 }-3x-6=0$, then the equation whose roots are ${ \alpha  }^{ 2 }+2$ and ${ \beta  }^{ 2 }+2$ will be

  1. $4{ x }^{ 2 }+49x-118=0$
  2. $4{ x }^{ 2 }-49x-118=0$
  3. $4{ x }^{ 2 }-49x+118=0$
  4. $4{ x }^{ 2 }+49x+118=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$2x^{2}-3x-6=0$

$\alpha+\beta=\dfrac{3}{2}$
$\alpha\beta=-3$
Now roots are $\alpha^{2}+2$   and   $\beta^{2}+2$
$sum=\alpha^{2}+2+\beta^{2}+2$
$=(\alpha+\beta)^{2}-2\alpha\beta+4$
$=\dfrac{9}{4}+6+4$
$=\dfrac{49}{4}$
$Product=(\alpha^{2}+2)(\beta^{2}+2)$
$=\alpha^{2}\beta^{2}+2(\alpha^{2}+\beta^{2})+4$
$=9+2\times\dfrac{33}{4}+4=13+\dfrac{33}{2}=\dfrac{59}{2}$
Equation
$x^{2}-\dfrac{49x}{4}+\dfrac{59}{2}=0$
$4x^{2}-49x+118=0$

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

If $\alpha, \beta$ are the roots of $x^2 + px+1=0$ and $\gamma, \delta $ are the roots of $x^2+qx+1=0$, then $(\alpha - \gamma) (\beta - \gamma)(\alpha - \delta) (\beta + \delta)=$

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

$\alpha, \beta$ are the roots of $x^2 + px+1=0$
$\Rightarrow \alpha+\beta = -p,  \alpha \beta =1$

$\gamma, \delta$ are the roots of $x^2+qx+1=0$
$\Rightarrow \gamma \delta =1, \gamma^2+q\gamma +1=0$,
$\delta^2 +q\delta +1=0$

$(\alpha - \gamma)(\beta -\gamma)(\alpha + \delta)(\beta + \delta)$
$=[\alpha \beta - \gamma (\alpha + \beta)+\gamma^2][\alpha \beta + \delta (\alpha + \beta) + \delta^2]$
$=(1+p \gamma + \gamma^2)(1-p\delta + \delta^2)$
$=(p \gamma - q \gamma)(-p \delta - q\delta)$
$=-\gamma \delta (p-q)(p+q)$
$=-(p^2-q^2) = q^2 - p^2$

Hence, option D.

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

Find the equation whose sum of roots and product of roots are the product and sum of roots of $x^2 + 5x + 6 = 0$ respectively.

  1. $x^2 - 6x - 5 = 0$
  2. $x^2 - 5x - 6 = 0$
  3. $x^2 + 11x - 1 = 0$

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In the given equation
sum of roots $= -5$ and product of roots $= 6$
The standard form of a quadratic equation is: $x^2 - (S)x + P = 0$, where S and P are sum and product of roots.
So according to question
$x^2 - (6)x + (-5) = 0$
The required equation will be $x^2 - 6x - 5 = 0$

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

If $\alpha, \beta $ are the roots of $ax^2+bx+c=0$ then the equation whose roots are $2+\alpha , 2+\beta$ is:

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

$\alpha, \beta$ are the roots of $\Rightarrow { ax }^{ 2 }+bx+c=0$
Then, $a(\alpha)^2 + b(\alpha)+c =0$
Now, $\alpha +2 = x $
Hence, $\alpha = x -2$
Thus, replace $\alpha$ by $x-2$ in the given equation,
Required equation is
$a(x-2)^2+b(x-2)+c=0$
$\Rightarrow a(x^2-4x+4)+bx-2b+c=0$
$\Rightarrow ax^2 +(b-4a) x+(4a-2b+c)=0$
$\Rightarrow ax^{ 2 }+x(b-4a)+4a-2b+c=0$

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

If $\alpha , \beta$ are the roots of the equation $x^2 - 3x + 1 = 0$, then the equation with roots $\displaystyle \frac{1}{\alpha - 2} , \frac{1}{\beta - 2}$ will be

  1. $x^2- x- 1 = 0$
  2. $x^2 + x - 1 = 0$
  3. $x^2 + x + 2 = 0$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\alpha , \beta$ are the roots of the equation $x^2 - 3x + 1 = 0$
$\Rightarrow  \alpha^2-3\alpha+1=0$ ------(1)
Let $\displaystyle\dfrac{1}{\alpha-2}=y$
$\Rightarrow\displaystyle \alpha=2+\dfrac{1}{y}$
From (1), we get
$\displaystyle\left(2+\dfrac{1}{y}\right)^2-3\left(2+\dfrac{1}{y}\right)+1=0$
$\Rightarrow\displaystyle \dfrac{(2y+1)^2}{y^2}-\dfrac{3(2y+1)}{y}+1=0$
$\Rightarrow y^2-y-1=0$
$\therefore$ The equation with roots $\displaystyle \dfrac{1}{\alpha - 2} , \dfrac{1}{\beta - 2}$ is $x^2-x-1=0$
Hence, option A.
Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $\alpha, \beta$ are roots of $ax^2+bx+c=0$, then one root of the equation $ax^2-bx(x-1) + c(x-1)^2=0$ is :

  1. $\displaystyle \left ( \frac{\alpha}{1- \alpha} \right )$
  2. $\displaystyle \left ( \frac{1-\beta}{\beta} \right )$
  3. $\displaystyle \left ( \frac{\alpha}{1+ \alpha} \right )$
  4. $\displaystyle \left ( \frac{\beta}{1+ \beta} \right )$
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

We have, $ax^2-bx^2+bx+cx^2-2cx+c=0$

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

Sum of the roots (S)
$\displaystyle \frac{b-2c}{a-b+c} = \frac{\left ( -\frac{b}{a} + \frac{2c}{a} \right )}{\left ( 1- \frac{b}{a} + \frac{c}{a}\right )}$

$\displaystyle S = \frac{\alpha+\beta+1\alpha \beta}{2+ \alpha + \beta+ \alpha \beta}=\frac{\alpha}{\alpha+1}+ \frac{\beta}{\beta+1}$

Product of the roots (P) $\displaystyle =\frac{c}{a-b+c}$

$\Rightarrow \displaystyle P= \frac{\left ( \frac{c}{a} \right )}{\left ( 1- \frac{b}{c}+\frac{c}{a}\right )}$

$\displaystyle=\frac{\alpha \beta}{1+\alpha+\beta+\alpha \beta} = \frac{\alpha}{(\alpha+1)} \cdot \frac{\beta}{(\beta+1)}$

Thus the roots are $ \displaystyle \frac{\alpha}{\alpha+1} and \frac{\beta}{\beta+1}$

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 $\beta$ be the roots of the equation $x^{2}+px+q = 0$, then the equation whose roots are $\alpha^{2}+\alpha\beta$ and $\beta^{2}+\alpha\beta$ is

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

Since $\alpha$ and $\beta$ are roots of the equation
$x^{2}+px+q = 0$, therefore
$\alpha + \beta = -p$      ...(i)
and $\alpha\beta = q$      ...(ii)
Sum of the roots $= \alpha^{2}+\alpha\beta+\beta^{2}+\alpha\beta$
$= (\alpha+\beta)^{2} = p^{2}$
Product of the roots $=(\alpha^{2}+\alpha\beta)(\beta^{2}+\alpha\beta)$
$= \alpha\beta (\alpha+\beta)^{2} = qp^{2}$
Required equation will be
$x^{2}$-(Sum  of  the  roots)$x$ + Product  of  the  roots = $0$
or $x^{2}-p^{2}x+qp^{2} = 0$

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

If $\alpha $ and $\beta \,\,\,\,$ are roots of equation $\,\,{x^3} - 2x + 3 = 0$,then the equation whose roots are $\,\dfrac{{\alpha  - 1}}{{\alpha  + 1}}$ and $\,\,\dfrac{{\beta  - 1}}{{\beta  + 1}}$ will be

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

$x^3-2x+3=0$

$\alpha+\beta=2$
$\alpha\beta=3$
$x^2-\left(\cfrac{\alpha-1}{\alpha+1}+\cfrac{\beta-1}{\beta+1}\right)x+\left(\cfrac{\alpha-1}{\alpha+1}\right)\left(\cfrac{\beta-1}{\beta+1}\right)=0$
$\Rightarrow x^2(\alpha+1)(\beta+1)-x((\alpha-1)(\beta+1)+(\beta-1)(\alpha+1))+(\alpha-1)(\beta-1)=0$
$\Rightarrow x^2(\alpha\beta+(\alpha+\beta)+1)-x(\alpha\beta+\alpha-\beta-1+\alpha\beta+\beta-\alpha-1)+(\alpha\beta-(\alpha+\beta)+1)=0$
$\Rightarrow x^2(3+2+1)-x(3-1)+(3-2+1)=0$
$\Rightarrow 6x^2-2x+2=0$
$\Rightarrow 3x^2-x+1=0$

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

Find a quadratic equation whose roots $\displaystyle \alpha$ and $ \displaystyle \beta $ are connected by the relation:
$\displaystyle \alpha +\beta = 2$ and $\displaystyle \frac{1-\alpha }{1+\beta }+\frac{1-\beta }{1+\alpha }= 2\left ( \frac{4\lambda ^{2}+15}{4\lambda ^{2}-1} \right )$

  1. $\displaystyle x^{2}-2x-\frac{\left ( 4\lambda ^{2}+11 \right )}{4}= 0$
  2. $\displaystyle x^{2}+2x-\frac{\left ( 4\lambda ^{2}-11 \right )}{4}= 0$
  3. $\displaystyle x^{2}-2x+\frac{\left ( -2\lambda ^{2}+11 \right )}{4}= 0$

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

$\displaystyle \alpha +\beta = 2$ and let $\displaystyle \alpha \beta = p$
$\displaystyle \therefore $ Equation is $\displaystyle x^{2}-2x+p= 0$ ...(1)
We have to find the value of p.
Now $\displaystyle

\frac{1-\alpha }{1+\beta }+\frac{1-\beta }{1+\alpha }= \frac{\left (

1-\alpha ^{2} \right )+\left ( 1-\beta ^{2} \right )}{1+\left ( \alpha

+\beta  \right )+p}$
or $\displaystyle \frac{2-\left ( \alpha

^{2}+\beta ^{2} \right )}{1+2+p}= \frac{2-\left { \left ( \alpha +\beta

 \right )^{2}-2\alpha \beta  \right }}{3+p}$
or $\displaystyle

\frac{2-4+2p}{3+p}:or:\frac{2\left ( p-1 \right )}{p+3}= 2\left (

\frac{4\lambda ^{2}+15}{4\lambda ^{2}-1} \right )$
or $\displaystyle \frac{p-1}{p+3}= \frac{4\lambda ^{2}+15}{4\lambda ^{2}-1}$
or $\displaystyle

p\left [ \left ( 4\lambda ^{2}-1 \right )-\left ( 4\lambda ^{2}+15

\right ) \right ]= 3\left ( 4\lambda ^{2}+15 \right )+\left ( 4\lambda

^{2}-1 \right )$
or $\displaystyle -16p= 16\lambda ^{2}+44= 4\left ( 4\lambda ^{2}+11 \right )$
$\displaystyle \therefore p= -\frac{\left ( 4\lambda ^{2}+11 \right )}{4}$
Putting for $p$ in (1) we get the required equation as
$\displaystyle x^{2}-2x-\frac{\left ( 4\lambda ^{2}+11 \right )}{4}= 0$

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

If $\alpha \neq \beta, \alpha^{2}=5\alpha -3$, and $\beta^{2}=5\beta-3$, then the equation having $\alpha/\beta$ and $\beta/\alpha$ as its roots is

  1. $3x^{2}-19x+3=0$
  2. $3x^{2}+19x-3=0$
  3. $3x^{2}-19x-3=0$
  4. $x^{2}+5x+3=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
${ \alpha  }^{ 2 }=5\alpha -3\quad { \beta  }^{ 2 }=5\beta -3$
Equation is ${ x }^{ 2 }-5x+3=0$
$\alpha +\beta =5\quad \alpha \beta =3$
If $\cfrac { \alpha  }{ \beta  } ,\cfrac { \beta  }{ \alpha  } $ are roots
Sum of roots$=\cfrac { \alpha  }{ \beta  } +\cfrac { \beta  }{ \alpha  } =\cfrac { { \alpha  }^{ 2 }+{ \beta  }^{ 2 } }{ \alpha \beta  } =\cfrac { { \left( \alpha +\beta  \right)  }^{ 2 }-2\alpha \beta  }{ \alpha \beta  } $
$=\cfrac { 25-2\left( 3 \right)  }{ 3 } =\cfrac { 19 }{ 3 } $
Products of roots$=\cfrac { \alpha  }{ \beta  } \times \cfrac { \beta  }{ \alpha  } =1$
${ x }^{ 2 }-\cfrac { 19 }{ 3 } x+1=0$
$\therefore 3{ x }^{ 2 }-19x+3=0$
Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

In a $\triangle ABC, C=90^{o}$. Then $\tan A$ and $\tan B$ are the roots of the equation

  1. $abx^{2}-c^{2}x+1=0$
  2. $abx^{2}-(a^{2}+b^{2})x+ab=0$
  3. $c^{2}x^{2}-abx+c^{2}=0$
  4. $ax^{2}-bx+a=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\tan A$ & $\tan B$ are roots 
Sum $=(\tan A+\tan B)$
Product $=\tan A\tan B$
$\angle C=90^o$
$\angle A+\angle B=90^o$
$\tan A \tan B=1$
$\tan B=1/\tan A$
$x^2-(\tan A+\tan B)x+1=0$
$x^2\dfrac {2}{\sin 2A}x+1=0$
$\Rightarrow \ x^2-\dfrac {c^2}{ab}x+1=0 \Rightarrow \ x^2 (ab)-c^2 (x)+1=0$
$\tan A+\tan B=\tan A+\dfrac {1}{\tan A}$
$=\dfrac {2\tan ^2 A+1}{2\tan A}=\dfrac {2}{\sin 2A}$
$\sin 2A=\dfrac {2\tan A}{1+\tan^2 A}$


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 $ are $\displaystyle \beta $ are the roots of $\displaystyle x^{2}+x+1=0$  then find the equation whose roots $\displaystyle \alpha ^{2}$ and $\displaystyle \beta ^{2}$

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

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

Product of roots $ = \alpha \times \beta =  \dfrac {1}{1} =1 $

Now, $ {\alpha}^{2} + \beta ^{2} = (\alpha + \beta )^{2} - 2(\alpha \times \beta)= (-1)^{2} - 2(1) = 1-2 = -1 $

And $ {\alpha}^{2} \times \beta ^{2} = (\alpha \times \beta )^{2} = 1 $

Equation whose roots are $ {\alpha}^{2} $ and $ \beta ^{2} $ is $ x^{2} -(Sum \ of \ roots)x +  Product \ of \ roots  = 0 $
$ => x^{2} -({\alpha}^{2} + \beta ^{2})x +  {\alpha}^{2} \times \beta ^{2}  = 0 $
$ => x^{2} -(-1)x+ 1  = 0 $
$ => x^{2} +x+ 1  = 0 $