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 the trapezium rule approximation errors and approximations the need for approximation

The positive root of ${x}^{2}-98.8=0$ after first approximation by Newton Raphson method assuming initial approximation to the root is $14$ is

  1. $9.821$
  2. $9.814$
  3. $9.715$
  4. $9.915$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here ${x} _{0}=14, f(x)={x}^{2}-78.8$
and $f'(x)=2x$
$\therefore$ ${x} _{1}={x} _{0}-\cfrac{f({x} _{0}}{f'({x} _{0})}$
$=14-\cfrac { { \left( 14 \right)  }^{ 2 }-\left( 78.8 \right)  }{ 2\times 14 } =9.814$

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

Using successive Bisection method find the second, third and fourth approximation of root of the  equation $x^3-3x-5=0$ in the interval $(2,2.5)$

  1. $ 2.375,2.135 \ \& \ \ 2.2815$
  2. $1.25,1.375 \ \ \& \ \ 1.4375$
  3. $4.23,3.214 \ \ \& \ \ 2.135$
  4. $2.4475,2.175 \ \ \& \ \ 3.2815$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We have to find the second,third and fourth approximation of root of the equation $x^3-3x-5=0$ in the interval $(2,2.5)$ using successive Bisection method.

$\textbf{Iteration 1: k=0}$

$c _0=\dfrac{a _0+b _0}{2}=\dfrac{2+2.5}{2}=2.25$

Since $f(c _0)f(a _0)=f(2.25)f(2)>0$

Therefore set $a _1=2.25,b _1=b _0$

$\textbf{Iteration 2: k=1}$

$c _1=\dfrac{a _1+b _1}{2}=\dfrac{2.25+2.5}{2}=2.375$

Since $f(c _1)f(a _1)=f(2.375)f(2.25)<0$

Therefore set $a _2=a _1,b _2=c _1$

$\textbf{Iteration 3: k=2}$

$c _2=\dfrac{a _2+b _2}{2}=\dfrac{2.25+2.375}{2}=2.3125$

Since $f(c _2)f(a _2)=f(2.3125)f(2.25)<0$

Therefore set $a _3=a _2,b _3=c _2$

$\textbf{Iteration 4: k=3}$

$c _3=\dfrac{a _3+b _3}{2}=\dfrac{2.25+2.3125}{2}=2.28125$

Thus the second,third and fourth approximations are $2.375,2.3125,2.28125$ respectively.

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The second and third approximation to the roots of $x^4-x-10=0$ in the interval $(1,2)$ is?

  1. $x _2=2.856,x _3=3.8561$
  2. $x _2=1.7756,x _3=1.061$
  3. $x _2=1.87409, x _3=1.85587$
  4. $x _2=7.856,x _3=1.8561$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Here, $x^4-x-10=0$


Let $f(x)=x^4-x-10$

$f'(x)=4x^3-1$

Here $f(1)=-10<0$ and $f(2)=4>0$

Root lies between $1$ and $2$

$x _0=\dfrac{1+2}{2}=1.5$

First Iteration:

$f(x _0)=f(1.5)=-6.4375$

$f'(x _0)=f'(1.5)=12.5$

$x _1=x _0-\dfrac{f(x _0)}{f'(x _0)}=1.5-\dfrac{-6.4375}{12.5}=2.015$

Second Iteration:


$f(x _1)=f(2.015)=4.47043$

$f'(x _1)=f'(2.015)=31.72541$

$x _2=x _1-\dfrac{f(x _1)}{f'(x _1)}=2.015-\dfrac{4.47043}{31.72541}=1.87409$


Third Iteration:


$f(x _2)=f(1.87409)=0.46155$

$f'(x _2)=f'(1.87409)=25.32882$

$x _3=x _2-\dfrac{f(x _2)}{f'(x _2)}=1.87409-\dfrac{0.46155}{25.32882}=1.85587$

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

Using successive Bisection method find the second, third and fourth approximation of root of the given  equation $x^3-x-4=0$ in the interval $(1,2)$

  1. $2.75,13.875 , 1.8125$
  2. $1.75,1.875 , 1.8125$
  3. $1.725,1.5 , 1.8125$
  4. $2.75,1.875 , 1.8125$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have to find the second,third and fourth approximation of root of the equation $x^3-x-4=0$ in the interval $(1,2)$ using successive Bisection method.

$\textbf{Iteration 1: k=0}$

$c _0=\dfrac{a _0+b _0}{2}=\dfrac{1+2}{2}=1.5$

Since $f(c _0)f(a _0)=f(1.5)f(1)>0$

Therefore set $a _1=1.5,b _1=b _0$

$\textbf{Iteration 2: k=1}$

$c _1=\dfrac{a _1+b _1}{2}=\dfrac{1.5+2}{2}=1.75$

Since $f(c _1)f(a _1)=f(1.75)f(1.5)>0$

Therefore set $a _2=c _1,b _2=b _1$

$\textbf{Iteration 3: k=2}$

$c _2=\dfrac{a _2+b _2}{2}=\dfrac{1.75+2}{2}=1.875$

Since $f(c _2)f(a _2)=f(1.875)f(1.75)<0$

Therefore set $a _3=a _2,b _3=c _2$

$\textbf{Iteration 4: k=3}$

$c _3=\dfrac{a _3+b _3}{2}=\dfrac{1.75+1.875}{2}=1.8125$

Thus the second,third and fourth approximations are $1.75,1.875,1.8125$ respectively.

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The second approximation of roots of $x^3-x-4=0$ in the interval $(1,2)$ by the method of false position is?

  1. $1.78049$
  2. $1.276$
  3. $2.123$
  4. $0.726$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Here, $f(x)=x^3-x-4=0$


first iteration:

]Here $f(1)=-4<0$ and $f(2)=2>0$

Now, root lies between $x _0=1$ and $x _1=2$

$x _2=x _0-f(x _0) \times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=1-(-4)\times \dfrac{2-1}{2-(-4)}=1.66667$

$f(x _2)=f(1.66667)=-1.03704<0$

2nd iteration:

Here $f(1.66667)=-1.03704<0$ and $f(2)=2>0$

Now, root lies between $x _0=1.66667$ and $x _1=2$

$x _3=x _0-f(x _0) \times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=1.67-(-1.04)\times \dfrac{2-1.67}{2-(-1.04)}=1.78049$

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

Using successive Bisection method find the second, third and fourth approximation of root of the equation $x^3+x^2-1$ in the interval $(0,1)$

  1. $0.75,1.875,0.8125$
  2. $1.75,0.875,0.8125$
  3. $0.75,0.875,0.8125$
  4. $0.75,0.875,1.8125$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We have to find the second,third and fourth approximation of root of the equation $x^3+x^2-1=0$ in the interval $(0,1)$ using successive Bisection method.

$\textbf{Iteration 1: k=0}$

$c _0=\dfrac{a _0+b _0}{2}=\dfrac{0+1}{2}=0.5$

Since $f(c _0)f(a _0)=f(0.5)f(0)>0$

Therefore set $a _1=0.5,b _1=b _0$

$\textbf{Iteration 2: k=1}$

$c _1=\dfrac{a _1+b _1}{2}=\dfrac{0.5+1}{2}=0.75$

Since $f(c _1)f(a _1)=f(0.75)f(0.5)>0$

Therefore set $a _2=c _1,b _2=b _1$

$\textbf{Iteration 3: k=2}$

$c _2=\dfrac{a _2+b _2}{2}=\dfrac{0.75+1}{2}=0.875$

Since $f(c _2)f(a _2)=f(0.875)f(0.75)<0$

Therefore set $a _3=a _2,b _3=c _2$

$\textbf{Iteration 4: k=3}$

$c _3=\dfrac{a _3+b _3}{2}=\dfrac{0.75+0.875}{2}=0.8125$

Thus the second,third and fourth approximations are $0.75,0.875,0.8125$ respectively.

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The third approximation of roots of $x^3-x-1=0$ in the interval $(1,2)$ by the method of false position is?

  1. $1.011$
  2. $2.265$
  3. $1.255$
  4. $1.294$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Here, $x^3-x-1=0$

Let $f(x)=x^3-x-1$
First Iteration:
Here, $f(1)=-1<0$ and $f(2)=5>0$
Now, Root lies between $x _0=1$ and $x _1=2$
$x _2=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=1-(-1)\times \dfrac{2-1}{5-(-1)}=1.16667$
Second Iteration:
Here, $f(1.16667)=-0.5787$ and $f(2)=5>0$

Now, Root lies between $x _0=1.16667$ and $x _1=2$
$x _3=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=1.17-(-0.58)\times \dfrac{2-1.17}{5-(-0.58)}=1.25311$

Third Iteration:

Here, $f(1.25311)=-0.28536$ and $f(2)=5>0$
Now, Root lies between $x _0=1.25311$ and $x _1=2$
$x _4=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=1.25-(-0.29)\times \dfrac{2-1.25}{5-(-0.29)}=1.29344$

Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

If $-9$ is a root of the equation $\begin{vmatrix} x & 3 & 7 \ 2 & x & 2 \ 7 & 6 & x \end{vmatrix}=0$, then the other two roots are

  1. $2,7$
  2. $-2,7$
  3. $2,-7$
  4. $-2,-7$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given $\begin{vmatrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{vmatrix}=0$
Simplifying matrix we get,
$x(x^2-12)-3(2x-14)+7(12-7x)=x^3-12x-6x+42+84-49x=x^3-67x+126$
The equation can be simplified by, $x^3-67x+126(x+9)(x^2-9x+14)$
$(x^2-9x+14)=(x^2-7x-2x+14)=(x-2)(x-7),x=2,7$
Hence the roots are $2,7,-9$.
Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

if $x= -5 $ is a root of $\displaystyle \Delta =\begin{vmatrix}
2x+1 & 4  & 8 \
2 & 2x  & 2 \
7 & 6  & 2x
\end{vmatrix}=0$ then the other two roots are

  1. $3 , 3.5$
  2. $1 , 3.5$
  3. $3 , 6$
  4. $2 , 6$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Applying $\displaystyle R _{1}\rightarrow R _{1}+R _{2}+R _{3}$ we get 

$\displaystyle \Delta =\begin{vmatrix}
2x+10 &2x+10  &2x+10 \ 
 2 & 2x  & 2 \ 
 7&6  & 2x
\end{vmatrix}$

Taking $2x+10$ common from $\displaystyle R _{1}$ and applying 

$\displaystyle C _{2}\rightarrow C _{2}-C _{1},C _{3}\rightarrow C _{3}-C _{1}$ we get

$\displaystyle \Delta =2\left ( x+5 \right )\begin{vmatrix}
1 &0  &0 \ 
 2 & 2x-2  & 0 \ 
 7&-1  & 2x-7
\end{vmatrix}$

$\displaystyle =2\left ( x+5 \right )\left ( 2x-2 \right )\left ( 2x-7 \right )$

Thus $\displaystyle \Delta =0 \ \Rightarrow x=-5,1,3.5$

$\displaystyle \therefore $ the other two roots are $1$ and $3.5$

Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

If $f(x) = ax^2 + bx + c, a, b, c \in  R$ and equation $f(x)- x = 0$ has non-real roots $\alpha, \beta$.  Let $\gamma, \delta$ be the roots of $f(f(x)) - x = 0$ ($\gamma, \delta$ are not equal to $\alpha, \beta$). Then $\begin{vmatrix} 2 & \alpha & \delta\ \beta & 0 & \alpha\ \gamma & \beta & 1\end{vmatrix} $ is

  1. 0

  2. purely real

  3. purely imaginary

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

$f(x) = ax^2 + bx + c; a, b, c \in  R$ and equation $f(x)- x = 0$ has imaginary roots $\alpha, \beta$ and $\gamma, \delta$ be the roots of $f(f(x)) - x = 0$
since $\alpha,\beta$ are roots of $f(x)- x = 0$
$f(\alpha)-\alpha =0$ $\Rightarrow f(\alpha)=\alpha$
$f(\beta)-\beta=0$ $\Rightarrow f(\beta)=\beta$
$f(f(\alpha))-\alpha = f(\alpha)-\alpha =0$
$f(f(\beta))-\beta = f(\beta)-\beta =0$
$\therefore  \alpha,\beta$ are also roots of $f(f(x)) - x = 0$ ------(*)
$f(x)- x= ax^2+(b-1)x+c=0$
roots are imaginary.
i.e $\alpha,\beta$ are conjugate to each other and $D<0$
$\Rightarrow (b-1)^2-4ac<0$ -------(1)
$f(f(x)) - x = a(ax^2+bx+c)^2+b(ax^2+bx+c)+c-x=0$
$\Rightarrow \left(ax^2+(b-1)x+c\right)\left(a^2x^2+(ab+a)x+ac+b+1\right)=0$
$D=(ab+a)^2-4a^2(ac+b+1) = a^2\left((b-1)^2-4ac\right)-4a^2<0$    ($\because$ from (1))
$\therefore \gamma,\delta$ are also imaginary roots and conjugate to each other.
$\begin{vmatrix} 2 & \alpha & \delta\ \beta & 0 & \alpha\ \gamma & \beta & 1\end{vmatrix} $ $= -3\alpha\beta+\alpha^2\gamma+\beta^2\delta$ ------ (2)
$\alpha\beta$ is real
$\alpha^2\gamma$ is conjugate to $\beta^2\delta$
$\Rightarrow \alpha^2\gamma+\beta^2\delta$ is real
from (2)
$\therefore \begin{vmatrix} 2 & \alpha & \delta\ \beta & 0 & \alpha\ \gamma & \beta & 1\end{vmatrix} $ is purely real.
Hence, option B.


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

Sum of roots is $-1$ and sum of their reciprocals is $\dfrac{1}{6}$, then equation is?

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

$\Rightarrow$  Let $\alpha$ and $\beta$ are roots of the equation.

According to the given condition,
$\Rightarrow$  $\alpha+\beta=-1$                             ------ ( 1 )
Again according to the given condition,
$\Rightarrow$  $\dfrac{1}{\alpha}+\dfrac{1}{\beta}=\dfrac{1}{6}$

$\Rightarrow$  $\dfrac{\beta+\alpha}{\alpha\beta}=\dfrac{1}{6}$

$\Rightarrow$  $6(\alpha+\beta)=\alpha\beta$
$\Rightarrow$  $6(-1)=\alpha\beta$                          [ From ( 1 ) ]
$\therefore$  $\alpha\beta=-6$               ----  ( 2 )
Now, required equation,
$\Rightarrow$  $x^2-(\alpha+\beta)x+(\alpha\beta)=0$
Using ( 1 ) and ( 2 ) we get,
$\Rightarrow$  $x^2-(-1)x+(-6)=0$
$\therefore$  $x^2+x-6=0$

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 $x^{3}-kx^{2}+14x-8=0$ are in geometric progression ,then $k=$

  1. -3

  2. 7

  3. 4

  4. 0
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $\dfrac{a}{r}.a.ar $ be the roots 

$\Rightarrow \dfrac{a}{r}.a.ar=8$

$\Rightarrow a^{3}=8$

$\Rightarrow a=2$

$a=2 $ is a root given equation

$\Rightarrow 8-4k+28-8=0$

$\Rightarrow K=7$

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

The quadratic equation whose roots are twice the roots of  $2 x ^ { 2 } - 5 x + 2 = 0$  is:

  1. $8 x ^ { 2 } - 10 x + 2 = 0$
  2. $x ^ { 2 } - 5 x + 4 = 0$
  3. $2 x ^ { 2 } - 5 x + 2 = 0$
  4. $x ^ { 2 } - 10 x + 6 = 0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l} Let\, \alpha \, and\, \beta \, be\, the\, root\, of\, the\, given\, equation. \ Then,\, \alpha +\beta =\frac { 5 }{ 2 } and\, \alpha \beta =\frac { 2 }{ 2 } =1 \ \therefore 2\alpha +2\beta  \ \therefore \left( { \alpha +\beta  } \right)  \ \therefore 5\left( { 2\alpha  } \right) \left( { 2\beta  } \right) =4 \ So\, the\, required\, equation\, is: \ { x^{ 2 } }-5x+4=0 \end{array}$


So, option $B$ is correct answer.

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

If $(b - c){x^2} + (c - a)x + (a - b) = 0$ has equal roots then $a,b,c$ are in :

  1. A.P.

  2. G.P.

  3. H.P.

  4. none
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
We have,

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

Comparing with the quadratic equation

$Ax^2+Bx+C=0$

$A=(b-c),B=c-a,C=a-b$

Discriminate when roots are equal

$D=B^2-4AC=0$

$D=(c-a)^2−4(b-c)(a-b)=0$

$D=(c^2+a^2−2ac)-4(ba-ac-b^2+bc)=0$

$D=c^2+a^2−2ac-4ab+4ac+4b^2-4bc=0$

$c^2+a^2+2ac-4b(a+c)+4b^2=0$

$(a+c)^2-4b(a+c)+4b^2=0$

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

$a+c=2b$

So,

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

Hence, this is the answer.
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 $ax^{2} \, + \, bx \, + \, c \, = \, 0$. The equation whose roots are as given below.
$\alpha \, + \,\dfrac{1}{\beta} \, , \, \beta \, + \, \dfrac{1}{\alpha}$ is $acx^2 \, + \, b(a \, + \, c) \, x \, + \, (a \, + \, c)^2 \, = \, 0$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\Rightarrow$  $\alpha$ and $\beta$ are roots of the equation $ax^2+bx+c=0$

$\Rightarrow$  $\alpha+\beta=\dfrac{-b}{a}$                       -------- ( 1 )
$\Rightarrow$  $\alpha\beta=\dfrac{c}{a}$                    ------- ( 2 )
Now,
$\Rightarrow$  $\alpha+\dfrac{1}{\beta}+\beta+\dfrac{1}{\alpha}=(\alpha+\beta)+\left(\dfrac{1}{\beta}+\dfrac{1}{\alpha}\right)$

                                    $=(\alpha+\beta)+\left(\dfrac{\alpha+\beta}{\alpha\beta}\right)$
                 
                                    $=\dfrac{-b}{a}+\dfrac{\dfrac{-b}{a}}{\dfrac{c}{a}}$     [ By using ( 1 ) and ( 2 ) ]

                                    $=\dfrac{-b}{a}-\dfrac{b}{c}$

                                    $=\dfrac{-bc-ba}{ac}$

$\therefore$   $\alpha+\dfrac{1}{\beta}+\beta+\dfrac{1}{\alpha}=\dfrac{-b(a+c)}{ac}$                    ----- ( 3 )

$\Rightarrow$  $\left(\alpha+\dfrac{1}{\beta}\right)\left(\beta+\dfrac{1}{\alpha}\right)=\alpha\beta+1+1+\dfrac{1}{\alpha\beta}$
 
                                            $=\dfrac{c}{a}+2+\dfrac{1}{\dfrac{c}{a}}$

                                            $=\dfrac{c}{a}+2+\dfrac{a}{c}$
 
                                            $=\dfrac{a^2+2ac+c^2}{ac}$

$\therefore$  $\left(\alpha+\dfrac{1}{\beta}\right)\left(\beta+\dfrac{1}{\alpha}\right)=\dfrac{a^2+2ac+c^2}{ac}$              ----- ( 4 )
Now, new equation,

$\Rightarrow$  $x^2-\left(\alpha+\dfrac{1}{\beta}+\beta+\dfrac{1}{\alpha}\right)x+\left[\left(\alpha+\dfrac{1}{\beta}\right)\left(\beta+\dfrac{1}{\alpha}\right)\right]=0$
By using ( 3 ) and ( 4 ),

$\Rightarrow$  $x^2+\dfrac{b(a+c)}{ac}x+\dfrac{a^2+2ac+c^2}{ac}$

$\Rightarrow$  $acx^2+b(a+c)x+(a+c)^2=0$