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 reciprocal equations theory of equations maths

If the reciprocal of every root of an equation is also a root of it, then the equation is said to be a

  1. reciprocal equation of first type

  2. reciprocal equation of second type

  3. reciprocal equation

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

An equation whose roots can be divided into pairs of numbers, each the reciprocal of the other or an equation which is unchanged if the variable is replaced by its reciprocal is known as reciprocal euation.

Multiple choice reciprocal equations theory of equations maths

The root of the reciprocal equation of second type and of odd degree is:

  1. $x=-1$
  2. $x=+1$
  3. $x=\pm1$
  4. $x=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$x = -1$ is a root of the reciprocal equation of first type and of odd degree. 
$x = 1$ is a root of the reciprocal equation of second type and of odd degree.
$x=±1$ are two roots of reciprocal equation of second type and of even degree. 
Multiple choice reciprocal equations theory of equations maths

lf $\mathrm{f}({x})=0$ is a reciprocal equation of second type and fifth degree, then a root of $\mathrm{f}({x})=0$  is:

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

When the reciprocal equation is of an odd degree, second type, $x=1$ is always a solution.

Multiple choice reciprocal equations theory of equations maths

The roots equation $x^4-3x^3+4x^2-3x+1=0$ is 

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

$x^4-3x^3+4x^2-3x+1=0$
This equation is resiprocal equation of first type as $a _{n-i}=a _i$
Dividing equation by $x^2$:
$x^2-3x+4-\dfrac3x+\dfrac{1}{x^2}=0$
$x^2+\dfrac{1}{x^2}-3x-\dfrac3x+4=0$
$\left(x+\dfrac1x\right)^2-2-3\left(x+\dfrac1x\right)+4=0$
$\left(x+\dfrac1x\right)^2-3\left(x+\dfrac1x\right)+2=0$
Let $x+\dfrac1x=y$
$y^2-3y+2=0$
$(y-1)(y-2)=0$
$y=1$ or $y=2$
For $y=1$:
 $x+\dfrac1x=1$
$x^2-x+1=0$
$D=(-1)^2-4(1)(1)=-3<0$
Hence no real value of x exists for this case.
For $y=2$:
$x+\dfrac1x=2$
$x^2-2x+1=0$
$(x-1)^2=0$
$x=1$
Hence solution of the given equation is x=1.

Multiple choice reciprocal equations theory of equations maths

An equation of the form $2x^4-3x^3+7x^2-3x+2=0$ is called a .................

  1. Reciprocal equation

  2. Radical equation

  3. Exponential equation

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

As 'The coefficients from beginning to end and vice versa are the same'
therefore given equation is a reciprocal equation.

Multiple choice reciprocal equations theory of equations maths

The roots of $a _ { 1 } x ^ { 2 } + b _ { 1 } x + c _ { 2 } = 0$ are reciprocal of the roots of the equation $a _ { 2 } x ^ { 2 } + b _ { 2 } x + c _ { 2 } = 0$

  1. $\dfrac { a _ { 1 } } { a _ { 2 } } = \dfrac { b _ { 1 } } { b _ { 2 } } = \dfrac { c _ { 1 } } { c _ { 2 } }$
  2. $\dfrac { b _ { 1 } } { b _ { 2 } } = \dfrac { c _ { 1 } } { a _ { 2 } } = \dfrac { a _ { 1 } } { c _ { 2 } }$
  3. $\dfrac { a _ { 1 } } { a _ { 2 } } = \dfrac { b _ { 1 } } { c _ { 2 } } = \dfrac { c _ { 1 } } { b _ { 2 } }$
  4. $a _ { 1 } = \dfrac { 1 } { a _ { 2 } } , b _ { 1 } = \dfrac { 1 } { b _ { 2 } } , c _ { 1 } = \dfrac { 1 } { c _ { 2 } }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given:${a} _{1}{x}^{2}+{b} _{1}x+{c} _{1}=0$     ........$(1)$
${a} _{2}{x}^{2}+{b} _{2}x+{c} _{2}=0$     ........$(2)$

Let $\alpha,\,\beta$ be the roots of ${a} _{1}{x}^{2}+{b} _{1}x+{c} _{1}=0$     ........$(1)$

$\Rightarrow\,\alpha+\beta=-\dfrac{{b} _{1}}{{a} _{1}}$ and 

$\alpha\beta=\dfrac{{c} _{1}}{{a} _{1}}$

Given:Roots of $(1)$ are reciprocal to $(2)$

$\dfrac{1}{\alpha}+\dfrac{1}{\beta}=-\dfrac{{b} _{2}}{{a} _{2}}$ and $\dfrac{1}{\alpha\beta}=\dfrac{{c} _{2}}{{a} _{2}}$

$\Rightarrow\,\dfrac{\alpha+\beta}{\alpha\beta}-\dfrac{{b} _{2}}{{a} _{2}}$ and $\dfrac{1}{\alpha\beta}=\dfrac{{c} _{2}}{{a} _{2}}$

Using $\alpha+\beta=-\dfrac{{b} _{1}}{{a} _{1}}$ and $\alpha\beta=\dfrac{{c} _{1}}{{a} _{1}}$ we have

$\Rightarrow\,\dfrac{-\dfrac{{b} _{1}}{{a} _{1}}}{\dfrac{{c} _{1}}{{a} _{1}}}=-\dfrac{{b} _{2}}{{a} _{2}}$ and $\dfrac{1}{\dfrac{{c} _{1}}{{a} _{1}}}=\dfrac{{c} _{2}}{{a} _{2}}$

$\Rightarrow\,\dfrac{-{b} _{1}}{{c} _{1}}=-\dfrac{{b} _{2}}{{a} _{2}}$ and
 
$\dfrac{{a} _{1}}{{c} _{1}}=\dfrac{{c} _{2}}{{a} _{2}}$

$\Rightarrow\,\dfrac{{b} _{1}}{{b} _{2}}=\dfrac{{c} _{1}}{{a} _{2}}$ and 

$\dfrac{{a} _{1}}{{c} _{1}}=\dfrac{{c} _{2}}{{a} _{2}}$

$\Rightarrow\,\dfrac{{b} _{1}}{{b} _{2}}=\dfrac{{c} _{1}}{{a} _{2}}$ and 

$\dfrac{{c} _{1}}{{a} _{1}}=\dfrac{{a} _{2}}{{c} _{2}}$

$\Rightarrow\,\dfrac{{b} _{1}}{{b} _{2}}=\dfrac{{c} _{1}}{{a} _{2}}$ and 

$\dfrac{{c} _{1}}{{a} _{2}}=\dfrac{{a} _{1}}{{c} _{2}}$

$\therefore\,\dfrac{{b} _{1}}{{b} _{2}}=\dfrac{{c} _{1}}{{a} _{2}}=\dfrac{{a} _{1}}{{c} _{2}}$

Option$(b)$ is correct.
Multiple choice business maths functions and graphs some functions and their graphs -i graphs of the form y=ax^2+bx+c introduction to sets

Let $f\left( x \right) = p{x^2} + qx - \left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right),\,\left( {p,q,a,b,c \in R} \right)(a,b,c$ are distinct). If both roots of $f(x)=0$ are non-real, then 

  1. $2\left( {p + q} \right) - \left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] > 0$
  2. $2\left( {p + q} \right) - \left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] < 0$
  3. $p - 2q - 2 - \left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] < 0$
  4. $p - 2q - 2 - \left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] > 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The expression (a-b)^2 + (b-c)^2 + (c-a)^2 is always positive for distinct a, b, c. If the roots of px^2 + qx - K = 0 are non-real, the discriminant q^2 + 4pK < 0. The options involve complex algebraic manipulations of these coefficients.

Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

If $a, b , c \in R $ and $3b^2 - 8ac < 0$ then the
equation $ax^4 + bx^3 +cx^2 +5x - 7=0$ has

  1. (a) all real roots

  2. (b) all imaginary roots

  3. (c) exactly two real and two imaginary roots

  4. (d) none
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a quartic equation ax^4 + bx^3 + cx^2 + dx + e = 0, the nature of roots depends on the discriminant. Given 3b^2 - 8ac < 0, the derivative of the function (a cubic) has only one real root, implying the quartic has no real roots or all real roots depending on the constant terms. However, standard analysis of this specific inequality often leads to the conclusion of all real roots in specific contexts.

Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

Condition for an irreducible quadratic equation is-

  1. discriminant is positive

  2. discriminant is negative

  3. discriminant is zero

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

Irreducible quadratic equation can not be reduced more i.e the quadratic equation which do not have real roots   

This means the roots are imaginary
So if roots are imaginary, then discriminant $D  <  0$
Hence, option B is correct.

Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

Number of real roots of equation 
(x+1) (x+2) (x+3) (x+4) -8 =0 is

  1. 0

  2. 2

  3. 4

  4. 3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\begin{matrix} \left( { x+1 } \right) \left( { x+2 } \right) \left( { x+4 } \right) =8 \ { x^{ 4 } }+{ 10^{ 3 } }+35{ x^{ 2 } }+50x+16=0 \ From\, \, Oescantes\, rule\, of\, sign\, of\, \, sign\,  \ There\, will\, be\, no\, positive\, \, roots\,  \ f\left( { -x } \right) =\, \, \, { x^{ 4 } }-10{ x^{ 3 } }+35{ x^{ 2 } }-50x+60=0 \ and\, posibility\, \, of\, negative\, roots\, \, and\, 0,2\, \, or\, \, 4 \ but\, no\, \, negative\, number\, making\, this\, equation\, '0'\, \, so\, it\, has\, no\, real\, roots\,  \  \end{matrix}$

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.