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.

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Equations and Roots Questions

Multiple choice maths equations and simple functions further equations solving linear equations with variable on both sides solution of linear equations in one variable

The number of solution(s) of the equation $[x]+2{-x}=3x$, is$/$are (where $[]$ represents the greatest integer function and ${ x}$ denotes the fractional part of x$)$:

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

Given; $[x]+2\left{-x\right}=3x$

$\left{-x\right}+[-x]=-x     (\because  \left{a\right}+[a]=a)$
$\therefore  \left{-x\right}=-x-[-x]$
We know that,  $[-x]=-1-[x]$
$[x]-2x-2[-x]=3x$
$[x]-2x+2+2[x]=3x$
$3[x]=5x-2$
L.H.S is integer
$\therefore$ R.H.S must be integer
$\therefore$  $x$must be integer
As $x$ is integer , $[x]=x$
$3x=5x-2$
$x=1$
Only one solution is possible.

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

The number of solution of $2\cos^2\dfrac{\pi}{2}\sin^2x=x^2+\dfrac{1}{x^2},\;0 \le x \le \dfrac{\pi}{2}$ is 

  1. Zero

  2. One

  3. Infinite many

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

$2\left ( \cos^2 \dfrac{\pi }{2} \right )\left ( \sin^{2}x \right )=x^{2}+\dfrac{1}{x^{2}}$

 
$=2(0) \sin^{2}x=x^{2}+\dfrac{1}{x^{2}}$ 

$=x^{2}+\dfrac{1}{x^{2}}=0$ 

not possible for any $ X\in R$ 

$\therefore $ no. of solutions = $0 $

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

The number of integers (positive, negative or zero) solutions of
$xy-6(x+y)=0$ with x is less than or equal to y is:

  1. 5

  2. 10

  3. 12

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

Rewrite as xy - 6x - 6y = 0, which factors to (x - 6)(y - 6) = 36. We need pairs (x-6, y-6) such that their product is 36 and x <= y. Possible pairs (u, v) where u*v=36 and u <= v: (-36, -1), (-18, -2), (-12, -3), (-9, -4), (-6, -6), (1, 36), (2, 18), (3, 12), (4, 9), (6, 6). There are 10 such pairs.

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

The solution of the equation ${\left| {x + 1} \right|^2} - \left| {x + 2} \right| - 26 = 0$ is:

  1. $ \dfrac{-1 + \sqrt{(109)}}{2}$,$ \dfrac{-3 - \sqrt{(101)}}{2}$
  2. $ - 7,\sqrt {29} $
  3. $ \pm \sqrt {29} $
  4. $ - 7,29$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$|x+1|^2$ will be always non negative so we can expand it and remove modulus.


So, Equation becomes $x^2+2x-25 -|x+2| = 0$ 


For $x > -2$
$x^2 + x - 27 = 0$ which gives $x=\dfrac{-1 ^+ _-\sqrt{(109)}}{2}$
But since we assumed $x>-2$, $x=\dfrac{-1 + \sqrt{109}}{2}$

Now for $x<-2$
$x^2+3x -23$ which gives x = $\dfrac{-3 ^+ _- \sqrt{101}}{2}$
But since we assumed $x<-2$, x = $\dfrac{-3-\sqrt{101}}{2}$

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

How many distinct real solutions does the equation $((x^2 - 2)^2 - 5)^2 = 1$ have ?

  1. 5

  2. 6

  3. 8

  4. 9
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This equation is equivalent to
$(x^2 - 2)^2 - 5 = 1$  or $(x^2 -2)^2 - 5 = -1$


The first is equivalent to $x^2 - 2 = \sqrt{6}$ or $\ x^2 - 2 = - \sqrt{6}$, 
$x=\pm (2+\sqrt 6)$ or $x^2 \neq -\sqrt 6 +2 $
with $2$ and $0$ solutions respectively (since - $\sqrt{6}$ + 2 < 0).
The latter is equivalent to $x^2 - 2 = 2$ or $x^2 - 2 = -2,$ 
$x=\pm 2$ or $x=0$
with $2$ and $1$ solution(s) respectively.
So we have $2+0+2+1 = 5$ solutions in total.

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

The number of ordered pairs of integers (x,y) satisfying the equation
${ x }^{ 2 }+6x+{ y }^{ 2 }=4$ is

  1. 2

  2. 4

  3. 6

  4. 8
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given,

$x^2+6x+y^2=4$ 

$\Rightarrow x^2+6x+9+y^2=4+9$

$\Rightarrow x^2+3^2+2(3x)+y^=13$

$\Rightarrow (x+3)^2+y^2=13$

sum of two squares is $13$

$\therefore $ when $(x+3)^2=9,x=0,-6$ and $y^2=2,-2$ $\Rightarrow 4$ ordered pairs.

$\therefore $ when $(x+3)^2=4,x=-1,-5$ and $y^2=3,-3$ $\Rightarrow 4$ ordered pairs.

A total of $8$ pairs.
Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

Number of solutions satisfying, $\sqrt { 5-{ log } _{ 2 }x } =3-{ log } _{ 2 }x$ are :

  1. 1

  2. 2

  3. 3

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

$\sqrt { 5-{ \log } _{ 2 }x } =3-{ \log } _{ 2 }x$                  ---- ( 1 )


Let $\log _2x=y$                     ----- ( 2 )


$\Rightarrow$  $\sqrt{5-y}=3-y$

Squaring both sides, 

$\Rightarrow$  $5-y=9-6y+y^2$

$\Rightarrow$  $y^2-5y+4=0$

$\Rightarrow$  $y^2-4y-y+4=0$

$\Rightarrow$  $y(y-4)-1(y-4)=0$

$\Rightarrow$  $(y-4)(y-1)=0$

$\Rightarrow$  $y=4$ and $y=1$

Substituting $y=1$ in ( 1 ) we get,

$\Rightarrow$  $\log _2x=1$

We know, $\log _ba=x\Rightarrow a=b^x$

$\therefore$  $x=2^1=2$

Substituting $y=4$ in ( 1 ) we get,

$\Rightarrow$  $\log _2x=4$

We know, $\log _ba=x\Rightarrow a=b^x$

$\therefore$  $x=2^4=16$

Substituting $\log _2x=1$ in ( 1 ) we get,

$\sqrt{5-1}=3-1$

$\Rightarrow$  $\sqrt{4}=2$

$\therefore$  $2=2$

Substituting $\log _2x=4$ in ( 1 ) we get,

$\sqrt{5-4}=3-4$

$\Rightarrow$  $\sqrt{1}=-1$

$\therefore$  $1=-1$

Hence, we can see only $\log _2x=1$ satisfying.

$\therefore$  $\sqrt { 5-{ \log } _{ 2 }x } =3-{ \log } _{ 2 }x$ has only $1$ solution.

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

$|x - 1| + |x + 3| + |x - 5| = k$
How many values does $k$ have.

  1. only one solution

  2. two solution

  3. no solution

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

The function f(x) = |x - 1| + |x + 3| + |x - 5| is a continuous, strictly increasing function for x > 5 and strictly decreasing for x < -3. It has a minimum value at x = 1. Depending on the value of k, there is either one solution, no solution, or a range of solutions.

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

Consider the equation $(1 + a + b)^{2} = 3(1 + a^{2} + b^{2})$, where a, b are real numbers.
Then

  1. there is no solution pair (a, b)

  2. there are infinitely many solution pairs (a, b)

  3. there are exactly two solution pairs (a, b)

  4. there is exactly one solution pair (a, b)
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$ (1+a+b)^{2} = 3(1+a^{2}+b^{2}) $

$ \Rightarrow (1+a^{2}+b^{2} + 2a + 2b + 2ab) = 3 + 3a^{2} +3b^{2} $
$ \Rightarrow a^{2}+b^{2} -ab-a-b+1 = 0 $
$ \Rightarrow (a-b)^{2} + ((a-1)(b-1)) = ((a-1)-(b-1))^{2} + ((a-1)(b-1)) = 0 $
Let $ a-1 = x $ and $ b-1 = y $

$ \Rightarrow (x-y)^{2} + (xy) = 0 $
$ \Rightarrow x^{2} - xy + y^{2} = 0 $                                 ...($1$)
Assume $ y \neq 0 $ and divide by $ y^{2} $

$ \Rightarrow \left ( \dfrac{x}{y} \right )^{2} - \dfrac{x}{y} + 1 = 0 $
Substitute $ t = \dfrac{x}{y} $
$ \Rightarrow t^{2} - t + 1 = 0 $
Discriminant $ D = 1-4 = -3 < 0 $. No solution here.

Assume $ y = 0 $ in equation ($1$). It satisfies the equation for $ x = 0 $. Hence the solution is $x=0$ and $y=0$

$ \Rightarrow a=1$ and $b=1$

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

The number of values of $k$ for which the system of equations 
$(k+1)x+8y = 4 $
$kx+(k+3)y = 3k-1$
has infinitely many solutions is

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

For infinitely many solutions,
$\displaystyle \frac { k+1 }{ k } =\frac { 8 }{ k+3 } =\frac { 4 }{ 3k-1 } $

$\Rightarrow k^{2}-4k+3=0 $ and $24k-8=4k+12$
$\Rightarrow k=3,k=1$ and $k=1$

Hence, $k=1$

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

Solve the equation for $x$.
$\left|\begin{matrix} a^2 & a & 1 \ \sin(n+1)x & \sin{nx} & \sin(n-1)x \ \cos(n+1)x & \cos{nx} & \cos(n-1)x\end{matrix}\right| = 0$. Given that $ a>0$

  1. $x = n\pi$
  2. $x = (n-1)\pi$
  3. $x = (n+1)\pi$

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

$\left|\begin{matrix} a^2 & a & 1 \ \sin(n+1)x & \sin{nx} & \sin(n-1)x \ \cos(n+1)x & \cos{nx} & \cos(n-1)x\end{matrix}\right| = 0$
 
$ \Rightarrow \displaystyle { a }^{ 2 }\left[ \sin { nx } .\cos { \left( n-1 \right) x } -\cos { nx } .\sin { \left( n-1 \right) x }  \right] +a\left[ \sin  (n-1)x.\cos  (n+1)x-\cos  (n-1)x.\sin  (n+1)x \right] \ +1\left[ \sin  (n+1)x.\cos { nx } -\cos  (n+1)x.\sin { nx }  \right] =0$

$ \displaystyle \Rightarrow a^{ 2 }\sin  x-a\sin  2x+\sin  x=0\ \displaystyle \Rightarrow \sin  x\left( { a }^{ 2 }+1-2a\cos { x }  \right) =0\ \displaystyle \Rightarrow \sin  x=0\; { or }\; \cos { x } =\frac { { a }^{ 2 }+1 }{ 2a } \ \displaystyle \Rightarrow \sin  x=0\; { or }\; \cos { x } =1\quad \left( \because { a }^{ 2 }+1\ge 2a,\; a>0\Rightarrow \frac { { a }^{ 2 }+1 }{ 2a } \ge 1 \right) \ \displaystyle \Rightarrow x=n\pi \; or\; x=2n\pi, \; n \in I \ \displaystyle\Rightarrow x=n\pi $

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

The number of solutions of the system of equations $2x+y-z=7   ,   x-3y-2z=1 ,  x+4y-3z=5,$ are 

  1. 0

  2. 1

  3. 2

  4. infinitely many
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
From given, we have,

$\Delta =\begin{vmatrix} 2 &1  &-1 \\ 1 & 3 &-2 \\  1& 4 &-3 \end{vmatrix}$

$=2(9-8)-1(-3-2)-1(4+3)$

$=0$

$\Delta _1=\begin{vmatrix} 7 &1  &-1 \\ 1 & 3 &-2 \\  5& 4 &-3 \end{vmatrix}$

$=7(9-8)-1(-3-10)-1(4+15)$

$=1\neq 0$

Hence, the given system has no solution.