Mathematics · Quantitative Aptitude

Equations and Roots

69 Questions

Equations and roots questions cover finding real solutions to algebraic, trigonometric, and polynomial equations. They are fundamental for advanced mathematics sections. Practicing these problems ensures quick recognition of underlying patterns and calculation shortcuts.

polynomial equationssystem of equationstrigonometric equationsabsolute value equations

Equations and Roots Questions

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

The number of distinct real roots of $\displaystyle \left | \begin{matrix}\sin x &\cos x  &\cos x \\cos x  &\sin x  &\cos x \\cos x  &\cos x  &\sin x \end{matrix} \right |= 0$ in the interval $\displaystyle -\frac{\pi }{4}\leq x\le\frac{\pi }{4}$ is

  1. 0

  2. 2

  3. 1

  4. 3
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$\displaystyle \left | \begin{matrix}\sin x &\cos x  &\cos x \\\cos x  &\sin x  &\cos x \\\cos x  &\cos x  &\sin x \end{matrix} \right |= 0$

Applying ${ C } _{ 1 }\rightarrow { C } _{ 1 }+{ C } _{ 2 }+{ C } _{ 3 }$, we get

$\begin{vmatrix} \sin { x } +2\cos { x }  & \cos { x }  & \cos { x }  \\ \sin { x } +2\cos { x }  & \sin { x }  & \cos { x }  \\ \sin { x } +2\cos { x }  & \cos { x }  & \sin { x }  \end{vmatrix}=0$

Taking $\left( \sin { x } +2\cos { x }  \right) $ common from ${C} _{1}$, we get

$ \Rightarrow \left( \sin { x } +2\cos { x }  \right) \begin{vmatrix} 1 & \cos { x }  & \cos { x }  \\ 1 & \sin { x }  & \cos { x }  \\ 1 & \cos { x }  & \sin { x }  \end{vmatrix}=0$

Applying ${R} _{2}\rightarrow{R} _{2}-{R} _{1};{R} _{3}\rightarrow{R} _{3}-{R} _{1}$

$\Rightarrow \left( \sin { x } +2\cos { x }  \right) \begin{vmatrix} 1 & \cos { x }  & \cos { x }  \\ 0 & \sin { x-\cos { x }  }  & 0 \\ 0 & 0 & \sin { x-\cos { x }  }  \end{vmatrix}=0\\ \Rightarrow \left( \sin { x } +2\cos { x }  \right) { \left( \sin { x } -\cos { x }  \right)  }^{ 2 }=0\\ \Rightarrow \sin { x } +2\cos { x } =0,\quad \sin { x } -\cos { x } =0\\ \Rightarrow \tan { x } =-2,\quad \tan { x } =1$

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

If a,b,c$\in $ R. Than the system of the equation is :$\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } -\frac { { z }^{ 2 } }{ { c }^{ 2 } } =1.\ \ \frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } +\frac { { z }^{ 2 } }{ { c }^{ 2 } } =1.\ \ \frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } -\frac { { z }^{ 2 } }{ { c }^{ 2 } } =1\ \ has\quad $.

  1. No solution

  2. a unique solution

  3. infinirty many solution

  4. finietil many solution
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

System of equations
$x + 2y + z = 0, 2x + 3y- z = 0 $ and $(tan\theta) x + y -3z = 0$ has non-trivial solution then number of value(s) of $\theta \epsilon (-\pi,\pi)$ is equal to?


  1. 0

  2. 1

  3. 2

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

For non-trivial solutions
$\begin{vmatrix}
1 &2  &1 \ 
2 &3  &-1 \ 
tan\theta &1  &-3 
\end{vmatrix}=0$
$\Rightarrow 6-5\tan\theta =0$
$\Rightarrow\displaystyle \tan \theta=\frac{6}{5}$
Hence, the number of solutions in $(-\pi,\pi)$ is 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

The number of real solution of $x-\dfrac{1}{x^2-4}=2-\dfrac{1}{x^2-4}$ is 

  1. $0$
  2. $1$
  3. $2$
  4. $infinitie$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given that:
$x-\dfrac{1}{x^2-4}=2-\dfrac{1}{x^2-4}$

$x^3-4x-1=2x^2-8-1$
$x^3-2x^2-4x+8=0$
$x^2(x-2)-4(x-2)=0$
$(x^2-4)(x-2)=0$
$(x-2)(x+2)(x-2)=0$
$(x-2)(x+2)=0$
$x=-2,+2$

Hence, 
There are two real solutions for the given expression.
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 equation
$\left| {\begin{array}{{20}{c}}  {{{\left( {1 + x} \right)}^2}}&{{{\left( {1 - x} \right)}^2}}&{ - \left( {2 + {x^2}} \right)} \   {2x + 1}&{3x}&{1 - 5x} \   {x + 1}&{2x}&{2 - 3x} \end{array}} \right| + \left| {\begin{array}{{20}{c}}  {{{\left( {1 + x} \right)}^2}}&{2x + 1}&{x + 1} \   {{{\left( {1 - x} \right)}^2}}&{3x}&{2x} \   {1 - 2x}&{3x - 2}&{2x - 3} \end{array}} \right| = 0$

  1. has no real solution
  2. fas $4$ real solutions

  3. has two real and two non-real solutions

  4. has infinite number of solutions, real or non-real
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sum of the two determinants is zero. By evaluating the determinants or checking for properties, one finds that the expression simplifies to a form that has no real solutions for x.

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

How many real solutions does the equation $x^{7}+14x^{5}+16x^{3}+30x-560=0$ has?

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

Let  $f(x) =x^{7}+14x^{5}+16x^{3}+30\mathrm{x} -560$

$\Rightarrow f'(x)=7x^{6}+70x^{4}+48x^{2}+30>0$  $\forall x \in R$

$\therefore f$ is increasing also $\displaystyle
\lim _{x\rightarrow\infty}f(x)=\infty$ ; $\displaystyle
\lim _{x\to-\infty}f(x)=-\infty$

Hence, $f(x) =0$ has exactly one real root .

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 sum of the solutions of the equation $64(81^{x})-84(144^{x})+27(256^{x})=0$  is:

  1. $1$
  2. $\dfrac{3}{2}$
  3. $\dfrac{5}{2}$

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

Given, $64(81^{ x })-84(144^{ x })+27(256^{ x })=0$

$ \Rightarrow 64\left( \left( \cfrac { 9 }{ 16 }  \right) ^{ x } \right) ^{ 2 }-84\left( \cfrac { 9 }{ 16 }  \right) ^{ x }+27=0$
Let $\left( \cfrac { 9 }{ 16 }  \right) ^{ x }=t$
$64t^{ 2 }-84t+27=0$
Solving this, we get
$\Rightarrow t=\cfrac { 3 }{ 4 } ,t=\cfrac { 9 }{ 16 } \Rightarrow x=\cfrac { 1 }{ 2 } ,x=1$
Hence, sum of roots is 
$\cfrac { 1 }{ 2 } +1=\cfrac { 3 }{ 2 } $
Hence, option 'B' is correct.

Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

Solutions of the equation $z^{7}-1=0$ are given by

  1. $\displaystyle z=-1,z=\cos \frac{2k\pi }{7}+i\sin \frac{2k\pi }{7},k=0,1,2, 3, 4, 5$
  2. $\displaystyle z=1\; and \; z=\cos \frac{2k\pi }{7}+i\sin \frac{2k\pi }{7},k=1,2,3, 4, 5, 6$
  3. $\displaystyle z=-1,z=\cos \frac{k\pi }{7}+i\sin \frac{k\pi }{7},k=0,1,2, 3, 4, 5$
  4. $\displaystyle z=1 \; and \; z=\cos \frac{k\pi }{7}+i\sin \frac{k\pi }{7},k=0,1,2,3, 4, 5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

${ z }^{ 7 }=1\ \Rightarrow z={ 1 }^{ \frac { 1 }{ 7 }  }={ \left( \cos { 0 } +i\sin { 0 }  \right)  }^{ \frac { 1 }{ 7 }  }$
$\Rightarrow z=\cos { \frac { 2k\pi  }{ 7 }  } +i\sin { \frac { 2k\pi  }{ 7 }  } $       ...{ De Moivre's Theorem}
Where $k=0,1,2,3,4,5,6$

For $k=0$
$z=1$
And $z=\cos { \frac { 2k\pi  }{ 7 }  } +i\sin { \frac { 2k\pi  }{ 7 }  } $ for $k=1,2,3,4,5,6$

Ans:B

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

The solution of the equation $(8)^{1+|cos x|+|cos x|^2+|cos x|^3+...)}=4^3$ in the interval $(-\pi, \pi)$ are.

  1. $\pm \dfrac {\pi }{3}, \pm \dfrac {\pi }{6}$
  2. $\pm \dfrac {\pi }{3}, \pm {\pi }$
  3. $\pm \dfrac {\pi }{3}, \pm \dfrac {2\pi }{3}$

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

 $(8)^{1+|cos x|+|cos x|^2+|cos x|^3+...)}=4^3$
$\Rightarrow (8)^{\dfrac{1}{1-|\cos x|}}=4^3=64=8^2$, since $|\cos x| < 1$ in $(-\pi, \pi)$
$\Rightarrow {\dfrac{1}{1-|\cos x|}}=2$
$\Rightarrow |\cos x|=\cfrac{1}{2}$
The solution in the given interval is,
$x=\pm \cfrac{\pi}{3}, \pm \cfrac{2\pi}{3}$
Hence, option 'C' is correct.

Multiple choice logarithm and its uses basic mathematical concepts physics

The equation ${ x }^{ \cfrac { 3 }{ 4 } { \left( \log _{ x }{ x }  \right)  }^{ 2 }+\log _{ x }{ x } -\cfrac { 5 }{ 4 }  }=\sqrt { 2 } $ has

  1. at least one real solution

  2. exactly three solutions

  3. exactly one irrational solution

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

The equation simplifies significantly because log_x(x) = 1. The equation becomes x^(3/4 * 1^2) + 1 - 5/4 = sqrt(2), which is x^(3/4) - 1/4 = sqrt(2). This has a real solution for x.

Multiple choice maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

The equation $x-\dfrac {8}{|x-3|}=3--\dfrac {8}{|x-3|}$ has

  1. Only one solution

  2. infinite solution

  3. no solution

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

Consider the following equation.

$ x-\dfrac{8}{|x-3|}=3-\dfrac{8}{|x-3|} $

$ x-\dfrac{8}{\pm \left( x-3 \right)}=3-\dfrac{8}{\pm \left( x-3 \right)} $

$ x-\dfrac{8}{\left( x-3 \right)}=3-\dfrac{8}{\left( x-3 \right)} $

$ x=3 $

 $ x+\dfrac{8}{\left( x-3 \right)}=3+\dfrac{8}{\left( x-3 \right)} $

$ x=3 $

Hence, this is the correct answer.