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

The condition for the equation $\displaystyle ax^{2}+bx+c= 0$ to have one root $n$ times the other, is:

  1. $\displaystyle na^{2}= bc\left ( n+1 \right )^{2}$
  2. $\displaystyle nb^{2}= ac\left ( n+1 \right )^{2}$
  3. $\displaystyle nb^{2}= ac\left ( n-1 \right )^{2}$

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

Given that the roots of the equation $ ax^2+bx+c=0 $ be such that one root is $n$ times the other. 
Let one root be $\alpha$, then the other root will be $n\alpha$ by given condition.
Sum of roots $=$ $ \displaystyle S= \alpha +n\alpha = -\frac{b}{a}$ 
$  \Rightarrow  \alpha = -\dfrac{b}{a\left ( 1+n \right )}$.....(1)
Product of roots $=  n\alpha ^{2}= \dfrac{c}{a}$ 
$ \Rightarrow  \alpha ^{2}= \dfrac{c}{an}$ ....(2)
From (1) and (2), we have
$ \Rightarrow   \dfrac{c}{an}= \dfrac{b^{2}}{a^{2}\left ( 1+n \right )^{2}} $
$ \Rightarrow  \displaystyle \therefore nb^{2}= ac\left ( n+1 \right )^{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

One root is three times the other, find the condition for a general quadratic equation

  1. $\displaystyle 3b^{2}= 16ac$
  2. $\displaystyle 3b^{2}= ac$
  3. $\displaystyle b^{2}= 16ac$
  4. $\displaystyle 9b^{2}= 16ac$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

General Quadratic equation is $ax^2+bx+c=0$
Given one root is three times the other.
i.e $\alpha,3\alpha$ are the roots.
Sum of the roots $=\displaystyle\frac{-b}{a}$
$\Rightarrow 4\alpha=\displaystyle\frac{-b}{a}$ ---(1)
Product of roots $=\displaystyle\frac{c}{a}$
$\Rightarrow 3\alpha^2=\displaystyle\frac{c}{a}$---(2)
From (1) and (2), we have
$3\left(\displaystyle\frac{-b}{4a}\right)^2=\displaystyle\frac{c}{a}$
$\therefore 3b^2=16ac$
Hence, option A 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

Roots of the equation $\displaystyle (x+1)(x+2)(x+2)(x+3)(x+6)=15x^{2}$ are

  1. all real & rational

  2. all non real

  3. two rational and two imaginary

  4. two imaginary and two irrational
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Rearranging the equation (x+1)(x+6) * (x+2)(x+3) = 15x^2 leads to (x^2 + 7x + 6)(x^2 + 5x + 6) = 15x^2. Dividing by x^2 gives (x + 6/x + 7)(x + 6/x + 5) = 15. Let y = x + 6/x. Then (y+7)(y+5) = 15, so y^2 + 12y + 20 = 0. Roots are y = -2, -10. Solving x + 6/x = -2 and x + 6/x = -10 yields two imaginary and two irrational 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 one root of $x^{3}+ax^{2}+bx+c=0$ is the sum of the other two roots, then

  1. $a^{3}=4(ab-c)$
  2. $a^{3}=4(ab-2c)$
  3. $a^{3}=ab-c$
  4. $a^{3}=ab-2c$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the roots be $\alpha,\beta,\gamma$
Then
$\alpha=\beta+\gamma$.
Hence
$\alpha+\beta+\gamma=-a$
$2(\beta+\gamma)=-a$
$\beta+\gamma=\alpha=\dfrac{-a}{2}$ ...(i)
$\alpha.\beta+\beta.\gamma+\gamma.\alpha=b$
$\alpha(\beta+\gamma)+\beta.\gamma=b$
$\alpha^{2}+\beta.\gamma=b$
Or 
$\dfrac{a^{2}}{4}+\beta.\gamma=b$
$a^{2}+4\beta.\gamma=4b$ ...(ii)
And 
$\alpha.\beta.\gamma=-c$
Or 
$\dfrac{-a}{2}.\beta.\gamma=-c$
Or 
$\beta.\gamma=\dfrac{2c}{a}$.
Then
$a^{2}+4\beta.\gamma=4b$
$a^{2}+4\dfrac{2c}{a}=4b$
$a^{3}+8c=4ab$
Or 
$a^{3}=4(ab-2c)$.

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

One root of $x^{3}+x^{2}-2x-1=0$ lies between 

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

$x^{ 3 }+x^{ 2 }-2x-1=0$
$f\left( 0 \right) =-1\ f\left( -1 \right) =1\ f\left( -2 \right) =-1\ f\left( -3 \right) =-13\ f\left( -4 \right) =-71$
As $f\left( -1 \right) >0$ and $f\left( -2 \right) <0$
Therefore one roots lies between $-2$ and $-1$

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 $\alpha,\beta$ of the equation $x^{4}-5x^{3}+11x^{2}-13x+6=0$ are connected by the relation $2\alpha+3\beta=7$, then the roots of the equation are

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

Let $\alpha ,\beta ,\gamma ,\delta $ are roots of $x^{ 4 }-5x^{ 3 }+11x^{ 2 }-13x+6=0$
${ s } _{ 1 }=\alpha +\beta +\gamma +\delta =5\ { s } _{ 4 }=\alpha \beta \gamma \delta =6$


For $\gamma ,\delta =1\pm i\sqrt { 2 } $ or $1\pm i\sqrt { 3 } \quad $
${ s } _{ 1 }\Rightarrow \alpha +\beta +2=5\Rightarrow \alpha +\beta =3$
Solving this with $2\alpha +3\beta =7$ we get
$\alpha =2$ and $\beta =1$

Now for $\gamma ,\delta =1\pm i\sqrt { 2 } $
$\alpha \beta \gamma \delta =2\left( 1+2 \right) =6$

And for $\gamma ,\delta =1\pm i\sqrt { 3 } $
$\alpha \beta \gamma \delta =2\left( 1+3 \right) =8$, not possible

Therefore, roots are $2, 1, 1\pm\sqrt{2}$
Hence, option 'C' 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

lf the difference of the squares of the roots of equation ${x}^{2} -6x+q=0$ is $24$, then the value of ${q}$ is:

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

Let $\alpha,\beta$ are roots of ${x}^{2}-6x+q=0,$ then

${ S } _{ 1 }=\alpha +\beta =6$
And ${ S } _{ 2 }=\alpha \beta =q$

Given ${ \alpha  }^{ 2 }-{ \beta  }^{ 2 }=24$
Now from ${ \left( \alpha -\beta  \right)  }^{ 2 }={ \left( \alpha +\beta  \right)  }^{ 2 }-4\alpha \beta $
$\Rightarrow { \left( \alpha -\beta  \right)  }^{ 2 }=36-4q\Rightarrow \left( \alpha -\beta  \right) =\sqrt { 36-4q } $

As ${ \alpha  }^{ 2 }-{ \beta  }^{ 2 }=24\Rightarrow \left( \alpha -\beta  \right) \left( \alpha +\beta  \right) =24$
$\Rightarrow \sqrt { 36-4q } \left( 6 \right) =24\Rightarrow \sqrt { 36-4q } =4$
$\Rightarrow 36-4q=16\Rightarrow 4q=20\Rightarrow q=5$

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 $\mathrm{a} _{\mathrm{n}}\mathrm{x}^{\mathrm{n}}+\mathrm{a} _{\mathrm{n}-1}\mathrm{x}^{\mathrm{n}-1}+\ldots\ldots+\mathrm{a} _{1}\mathrm{x}=0,\ \mathrm{a} _{1}\neq 0,\ \mathrm{n}\geq 2$, has a positive root $\mathrm{x}=\alpha$, then the equation $\mathrm{n}\mathrm{a} _{\mathrm{n}}\mathrm{x}^{\mathrm{n}-1}+(\mathrm{n}-1)\mathrm{a} _{\mathrm{n}-1}\mathrm{x}^{\mathrm{n}-2}+\ldots..+\mathrm{a} _{1}=0$ has a positive root, which is 

  1. greater than $\alpha$
  2. smaller than $\alpha$
  3. greater than or equal to $\alpha$
  4. equal to $\alpha$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

=$ \because { a } _{ n }{ x }^{ 2\  }+{ a } _{ n }+{ x }^{ n-1 }+............+{ a } _{ 1 }x=\quad 0\quad \quad \quad { a } _{ 1 }\neq 0\quad n\ge 2 $

= has the root $x=\infty$ 
= ${ f }^{ 1 }(x)=\quad x{ a } _{ n }{ x }^{ n-1 }+\quad (x-1)\quad { a } _{ n-1 }{ x }^{ n-2 }+.......{ a } _{ n }$
= $\because f(x)=0$
Let us take an example to see 
Let a quadratic equation ${ x }^{ 2 }+2x-3=0$
${ x }^{ 2 }+3x-x-3=0$
$x(x+3)-1(x+3)=0 ........(i)$
$x=1\quad x=-3$
Now ${ f }^{ 1 }(x)=\quad 2x+1$
${ f }^{ 1 }(x)=\quad 0\quad =>\quad x=\quad -\cfrac { 1 }{ 2 } ..........(ii) $
From (i) and (ii) we can see that
The root of ${ f }^{ 1 }(x)$ is always less than the root of $f(x)$
Hence we can conclude
for $n{ a } _{ n }{ x }^{ n-1 }+(n-1){ a } _{ n-1 }{ x }^{ n-2 }+......{ a } _{ 1 }$
has roots always less than $\alpha $ for the value of $\alpha$.


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 $x^{3}+ax+b=0$ is zero, then the value of $b$, is:

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

$let\quad \alpha ,-\alpha ,\beta \quad be\quad the\quad roots\ Given\quad sum\quad of\quad the\quad roots\quad is\quad zero\ \alpha -\alpha +\beta =0\ \beta =0\ Therefore\quad product\quad of\quad the\quad roots\quad is\quad zero,\quad i.e.,\quad b=0$

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 $\mathrm{x}^{2}-\mathrm{x}-\mathrm{k}=0(\mathrm{k}>0)$ is the square of the other root, then $\mathrm{k}=$ 

  1. $ 2\pm\sqrt{5}$
  2. $ 2+\sqrt{5}$
  3. $ 2-\sqrt{5}$
  4. $1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$ Let\quad \alpha \quad and\quad \alpha ^{ 2 }\quad be\quad the\quad roots\quad as\quad per\quad the\quad given\quad condition.\ \therefore \quad \alpha +\alpha ^{ 2 }=1\quad and\quad { \alpha  }^{ 3 }=-k\ Now\quad (\alpha +\alpha ^{ 2 })^{ 3 }=1\ \Rightarrow { \alpha  }^{ 3 }+({ \alpha  }^{ 2 })^{ 3 }+3{ \alpha  }^{ 3 }(\alpha +\alpha ^{ 2 })=1\ Replacing\quad { \alpha  }^{ 3 }\quad by\quad -k\quad we\quad get\quad \ -k+k^{ 2 }-3k-1=0\ \Rightarrow { k }^{ 2 }-4k-1=0\ \Rightarrow k=\frac { 4+\sqrt { 20 }  }{ 2 } =2+\sqrt { 5 } \ or\quad k=\frac { 4-\sqrt { 20 }  }{ 2 } =2-\sqrt { 5 } <0\ But\quad k>0\quad therefore\quad we\quad reject\quad this\quad value.\ \therefore \quad k=2+\sqrt { 5 } \ Answer-\quad Option\quad B. $

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 sum of the roots of the equation $ax^2+bx+c=0$ is equal to sum of their squares, then

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

Let $\alpha, \beta$ are roots of $\displaystyle a{ x }^{ 2 }+bx+c=0$, then


$\displaystyle \alpha +\beta =-\frac { b }{ a } $

$\displaystyle \alpha \beta =\frac { c }{ a } $

As sum of roots is equal to sum of their square, then 
$\displaystyle \alpha +\beta ={ \alpha  }^{ 2 }+{ \beta  }^{ 2 }$

$\displaystyle \Rightarrow \alpha +\beta ={ \left( \alpha +\beta  \right)  }^{ 2 }-2\alpha \beta $

$\displaystyle \Rightarrow -\frac { b }{ a } ={ \left( -\frac { b }{ a }  \right)  }^{ 2 }-2\left( \frac { c }{ a }  \right) $

$\displaystyle \Rightarrow -\frac { b }{ a } =\frac { { b }^{ 2 } }{ { a }^{ 2 } } -\frac { 2c }{ a } $

$\displaystyle \Rightarrow -ab={ b }^{ 2 }-2ac$

$\displaystyle \Rightarrow { b }^{ 2 }+ab-2ac=0$ 

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 sum of the squares of the roots of $x^{2}+px-3=0$ is $10$, then $p=$

  1. $ \pm 2$
  2. $\pm 3$
  3. $ 5$
  4. $-5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $\alpha,\beta$ are roots of ${x}^{2}+px-3=0,$ then

${ S } _{ 1 }=\alpha +\beta =-p$
And ${ S } _{ 2 }=\alpha \beta =-3$

Given ${ \alpha  }^{ 2 }+{ \beta  }^{ 2 }=10$
Now from ${ \left( \alpha +\beta  \right)  }^{ 2 }={ \alpha  }^{ 2 }+{ \beta  }^{ 2 }+2\alpha \beta $
$\Rightarrow { \left( -p \right)  }^{ 2 }=10+2\left( -3 \right) =10-6=4$
$\Rightarrow { p }^{ 2 }=4$
$\Rightarrow p=\pm 2$

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

If the sum of two roots of the equation  $x^{4}-x^{3}+2x^{2}+kx+17=0$ equals to the sum of the other two, then $k $ is equal to

  1. $\displaystyle \frac{7}{8}$
  2. $-\displaystyle \frac{7}{8}$
  3. $\displaystyle \frac{9}{8}$
  4. $-\displaystyle \frac{9}{8}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $\alpha ,\beta ,\gamma ,\delta $ be the roots of ${ x }^{ 4 }-{ x }^{ 3 }+2{ x }^{ 2 }+kx+17=0$
Such that $\alpha +\beta =\gamma +\delta $
Then ${ s } _{ 1 }=\alpha +\beta +\gamma +\delta =1\Rightarrow \alpha +\beta =\cfrac { 1 }{ 2 } $ 
${ s } _{ 2 }=\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta =2\Rightarrow { \left( \alpha +\beta  \right)  }^{ 2 }+\alpha \beta +\gamma \delta =2$
$\Rightarrow \alpha \beta +\gamma \delta =2-\cfrac { 1 }{ 4 } =\cfrac { 7 }{ 4 } $   ...(1)
${ s } _{ 3 }=\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta =-k\Rightarrow \left( \alpha +\beta  \right) \left( \alpha \beta +\gamma \delta  \right) =-k$
$\Rightarrow \left( \alpha \beta +\gamma \delta  \right) =-2k$   ...(2)
From (1) and (2), we have
$-2k=\cfrac { 7 }{ 4 } \Rightarrow k=-\cfrac { 7 }{ 8 } $

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 $P(x) = x^{32} - x^{25} + x^{18} - x^{11} + x^{4} - x^{3} + 1$. Which of the following are CORRECT?

  1. Number of real roots of $P(x) = 0$ are zero
  2. Number of imaginary roots of $P(x) = 0$ are $32$
  3. Number of negative roots of $P(x) = 0$ are zero
  4. Number of imaginary roots of $P(x) + P(-x) = 0$ are $32$
Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
Explanation

P(x) = x^32 - x^25 + x^18 - x^11 + x^4 - x^3 + 1. For x >= 1, P(x) > 0. For 0 <= x < 1, P(x) > 0. For x < 0, let x = -y (y > 0), P(-y) = y^32 + y^25 + y^18 + y^11 + y^4 + y^3 + 1 > 0. Thus, there are no real roots. Since there are no real roots, all 32 roots must be imaginary. P(-x) also has no real roots, so its roots are also imaginary.