Quantitative Aptitude

Algebra

3,715 Questions
All Polynomial Roots and Coefficients 2451 Zero-Product Root Finding 667 Linear System Manipulation 162 Polynomial Theorems and Reduction 85 Algebraic Identity Simplification 82 Surds and Indices 70 Modulus and Absolute Value 54 Algebraic Expression Substitution 46 Algebraic Fraction Simplification 33 Logarithmic Equations 25 Repeated Series Sum 12 Linear Equation Word Problems 9

Algebra Questions

Multiple choice reciprocal equations theory of equations maths

Determine the root of the equation: $\dfrac{9}{x}-\dfrac{7}{x}=1$

  1. $x=2$
  2. $x=-2$
  3. $x=1$

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

Given reciprocal equation can be written as

$\dfrac{9}{x}=\dfrac{7+x}{x}$
Cancelling out the denominator on both side, we get
$9=7+x$
$\Rightarrow x=2$
Hence, option A is correct.

Multiple choice reciprocal equations theory of equations maths

Solve the equation: $x^{-2}-2x^{-1}=8$

  1. $\dfrac{3}{4}, \dfrac{-1}{2}$
  2. $\dfrac{1}{4}, \dfrac{-1}{3}$
  3. $\dfrac{1}{3}, \dfrac{-1}{2}$
  4. $\dfrac{1}{4}, \dfrac{-1}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given, $x^{-2}-2x^{-1}=8$

$\Rightarrow \dfrac {1}{x^2}-\dfrac {2}{x}=8$
$\Rightarrow \dfrac {1-2x}{x^2}=8$
$\Rightarrow 1-2x=8x^2$
$\Rightarrow 8x^2+2x-1=0$
$\Rightarrow (2x+1)(4x-1)$
$\Rightarrow x=\dfrac {1}{4}, \dfrac {-1}{2}$

Multiple choice reciprocal equations theory of equations maths

Solve the equation $\sqrt{4x^2-7x-15}-\sqrt{x^2-3x}=\sqrt{x^2-9}$

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

Given equation is $\sqrt { 4{ x }^{ 2 }-7x-15 } =\sqrt { { x }^{ 2 }-9 } +\sqrt { { x }^{ 2 }-3x } $

$\Rightarrow \sqrt { x-3 } (\sqrt { 4x+5 } )=\sqrt { x-3 } (\sqrt { x+3 } +\sqrt { x } )$
Therefore $x=3$ is one solution and $\sqrt { 4x+5 } =\sqrt { x+3 } +\sqrt { x } $
By squaring above equation on both sides , we get $x+1=\sqrt{x(x+3)}$
Again square it on both sides , we get $x^{2}+2x+1=x^{2}+3x$
$\Rightarrow x=1$
Therefore option $D$ is correct

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 linear programming problems structure of linear programming model linear programming problem operations research

If $l,m,n$ be three positive roots of the equation $x^3-ax^2+bx+48=0$, then the minimum value of $\dfrac 1l +\dfrac 2m+\dfrac 3n$ is

  1. $1$
  2. $2$
  3. $\dfrac {-3}{2}$
  4. $\dfrac 52$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

we Know that, $A.M.\geq G.M.$

$\implies \dfrac{a+b+c}{3}\geq \sqrt[3]{abc}$

let $a=\dfrac{1}{l}, b=\dfrac{2}{m}, c=\dfrac{3}{n}$

Therefore,

$\dfrac{1}{3}(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq \sqrt[3]{(\dfrac{1\times2\times3}{lmn})}$


$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq 3\times\sqrt[3]{(\dfrac{1\times2\times3}{lmn})}$

Given, the roots of the polynomial $x^3-ax^2+bx+48=0$ are $l,m,n$
Therefore, the product of the roots $lmn=-(\dfrac{48}{1})=-48$

Substituting $lmn=-48$ in the above equation

$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq 3\times\sqrt[3]{(\dfrac{6}{-48})}$

$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq 3\times\sqrt[3]{(\dfrac{1}{-8})}$

$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq 3\times\sqrt[3]{(-\dfrac{1}{2})^3}$

$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq 3\times(-\dfrac{1}{2})$

$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq (-\dfrac{3}{2})$

therefore, the minimum value is $-\dfrac{3}{2}$

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

Is the following quadratic polynomial reducible or irreducible?
$f(x) = -2x^2-2x-1$

  1. Reducible with one real root

  2. Reducible with two real roots

  3. Irreducible

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

To check whether the given quadratic polynomial is reducible or irreducible, we need to calculate the discriminant

Calculate the discriminant for the equation, $-2x^2-2x-1=0$

$D=b^{2}-4ac=(-2)^2-4(-2)(-1)=-4<0$
Quadratic equation is irreducible if $D<0$
$\therefore$ The quadratic polynomial is irreducible.

Correct option is C

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

Which of the following quadratics is irreducible?

  1. $2x^2 - 5x + 3$
  2. $2x^2 - 5x - 3$
  3. $5x^2 - 2x + 3$
  4. $5x^2 - 2x - 3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We can check by comparing the D of quadratic equations with their relation with 0


$2x^2-5x+3=0$
$D=(-5)^2-4(2)(3)=25-24=1>0$
Reducible

$2x^2-5x-3=0$
$D=(-5)^2-4(2)(-3)=25+24=49>0$
Reducible

$5x^2-2x+3=0$
$D=(-2)^2-4(5)(3)=4-60=-56<0$
Irreducible


$5x^2-2x-3=0$
$D=(-2)^2-4(5)(-3)=4+60=64>0$
Rreducible


Therefore correct option is C


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 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 $f(x) = ax^2 + bx + c, a, b, c \in  R$ 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

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

The sum and the product of the zeroes of a quadratic polynomial are $ \dfrac{-1}{2} $ and $ \dfrac{1}{2}$ respectively, then the polynomial is :

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

Given: Sum of zeroes $=-\dfrac 12$ and product of zeroes $=\dfrac 12$

We know,
$x^2-(\text{sum of zeroes})x+(\text{product of zeroes})=0$
$\Rightarrow x^2-\left(-\dfrac 12\right)x+\dfrac 12=0$
$\Rightarrow 2x^2+x+1=0$
is the required polynomial.

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

The sum and the product of zeroes of a quadratic polynomial $p(x)$ are $-7$ and $-10$ respectively. Then $p(x)$ is :

  1. $x^{2}-7x-10$
  2. $x^{2}-7x+10$
  3. $x^{2}+7x-10$
  4. $x^{2}+7x+10$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given: Sum of zeroes $=-7$ and product of zeroes $=-10$
We know that
$p(x)=x^2-(\text{sum of zeroes})x+(\text{product of zeroes})$
$\Rightarrow p(x)=x^2-(-7)x+(-10)$
$\Rightarrow p(x)=x^2+7x-10$
is the required polynomial.