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 maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

If $\alpha$ and $\beta$ are the roots of $ax^2+bx+c=0$, then the quadratic equation whose roots are $\cfrac{1}{\alpha}$  and $\cfrac{1}{\beta}$ is

  1. $ax^2+bx+c=0$
  2. $bx^2+ax+c=0$
  3. $cx^2+bx+a=0$
  4. $cx^2+ax+c=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
The quadratic equation whose roots are  $\dfrac{1}{\alpha }$ & $\dfrac{1}{\beta }$      is  $x^{2}-\left(\dfrac{1}{\alpha }+\dfrac{1}{\beta }\right)x+\dfrac{1}{\alpha \beta }$
 
=> $ x^{2}-\left(\dfrac{\alpha + \beta }{\alpha \beta}\right)x+\dfrac{1}{\alpha \beta } = 0$
 
also we know that $\alpha +\beta =\dfrac{-b}{a}$ and $\alpha \beta =\dfrac{c}{a}$ as $\alpha,\beta$ are roots of the equation $ax^2+bx+c$
 
=> $ x^{2}-\left(\dfrac{-b }{c}\right)x+\dfrac{a}{c } = 0$
 
=> $ x^{2}+\left(\dfrac{b }{c}\right)x+\dfrac{a}{c } = 0$
 
=>  $  cx^{2}+b x+a = 0 $


Multiple choice maths linear graphs quadrants the cartesian plane the cartesian system

The abscissa of two points A and B are the roots of the equation ${x^2} + 2ax - {b^2}$ and their ordinates are the root of the equation ${x^2} + 2px - {q^2}=0$. the equation of the circle with AB as diameter is 

  1. ${x^2} + {y^2} + 2ax + 2py + {b^2} + {q^2} = 0$
  2. ${x^2} + {y^2} - 2ax - 2py - {b^2} - {q^2} = 0$
  3. ${x^2} + {y^2} + 2ax + 2py - {b^2} - {q^2} = 0$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
${ x }^{ 2 }+2ax-{ b }^{ 2 }=0$

Let roots be ${ x } _{ 1 }$ & ${ x } _{ 2 }$

${ x } _{ 1 }+{ x } _{ 2 }=\dfrac{-b}{a}=-2a$

$ { x } _{ 1 }{ x } _{ 2 }=\dfrac{c}{a}=-{ b }^{ 2 }$

$ { x }^{ 2 }+2px-{ q }^{ 2 }=0$

Let roots be ${ y } _{ 1 }$ & ${y } _{ 2 }$

${ y } _{ 1 }+{ y } _{ 2 }=-2p$

$ { y} _{ 1 }{ y } _{ 2 }=-{ q }^{ 2 }$

Equation in diametric form is 
${ x }^{ 2 }+{ y }^{ 2 }-({ x } _{ 1 }+{ x } _{ 2 })x-({ y } _{ 1 }+{ y } _{ 2 })y+{ x } _{ 1 }{ x } _{ 2 }+{ y } _{ 1 }{ y } _{ 2 }=0$

$ { x }^{ 2 }+{ y }^{ 2 }+2ax+2py-{ b }^{ 2 }-{ q }^{ 2 }=0$
Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

If the roots of the equation $x^2 - 4x + 1 = 0$ are the lengths of the semi-major axis and semi-minor axis of an ellipse, then the eccentricity of the ellipse lies between

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

Roots of x^2 - 4x + 1 = 0 are 2 +/- sqrt(3). Thus a = 2 + sqrt(3) and b = 2 - sqrt(3). e^2 = 1 - b^2/a^2. Calculation shows e is between 1/3 and 1/2.

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

If $\alpha, \beta$ are the roots of the equation $x^2 - 3x + 1=0$, then the equation with roots $\displaystyle \frac{1}{\alpha - 2}, \frac{1}{\beta - 2}$ will be-

  1. $x^2 - x - 1 = 0$
  2. $x^2 + x - 1 = 0$
  3. $x^2 + x + 2 = 0$

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

Let y = 1/(x-2). Then x-2 = 1/y, so x = 2 + 1/y = (2y+1)/y. Substitute into x^2 - 3x + 1 = 0: ((2y+1)/y)^2 - 3((2y+1)/y) + 1 = 0. Simplifying leads to y^2 - y - 1 = 0.

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

The number of roots of the equation  $\displaystyle x-\frac{2}{(x-1)}=1-\frac{2}{(x-1)}$ is 

  1. 0

  2. 1

  3. 2

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

Given, $ x - \cfrac {2}{(x-1)} = 1 - \cfrac {2}{(x-1)} $

Cancelling out $ - \cfrac {2}{(x-1)} $ from LHS and RHS we get, $ x = 1 $
But when $ x = 1 $, denominator of fraction $ - \cfrac {2}{(x-1)} $ is $ 0 $, which is not defined.
Hence, there is no root of this equation.

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

If the roots of the equation $\displaystyle px^{2}+qx+r=0$ are in the ratio $\displaystyle \varphi \ : \ m,$ then 

  1. $\displaystyle (\varphi +m)^{2}qp=\varphi mr^{2}$
  2. $\displaystyle (\varphi +m)^{2}pr=\varphi mq$
  3. $\displaystyle (\varphi +m)^{2}pr=\varphi mq^{2}$

  4. None of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let the roots be $\varphi r$ and $mr$
Sum of roots $= \varphi r+mr=-\dfrac{q}{p}$
$\therefore  (\varphi+m)r=-\dfrac{q}{p}$
$\therefore (\varphi+m)^2r^2=\dfrac{q^2}{p^2}$   ...(1)

Product of roots $= (\varphi r)(mr)=\dfrac{r}{p}$
$\therefore  \varphi mr^2=\dfrac{r}{p}$    ...(2)

Dividing equation (1) by (2), we get

$\dfrac{(\varphi+m)^2r^2}{\varphi mr^2}=\dfrac{\frac{q^2}{p^2}}{\frac{r}{p}}$

$\therefore \dfrac{(\varphi+m)^2}{\varphi m}=\dfrac{q^2}{pr}$

$\therefore (\varphi+m)^2pr=\varphi mq^2$
Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

For the equation $|x|^{2}+|x|-6=0$, the roots are

  1. one and only one real number.

  2. real with sum one.

  3. real with sum zero.

  4. real with product zero.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For x>0 equation is $x^2+x-6=0$
$(x-2)(x+3)=0$
$x=2$
$x$ cant be equal to -3 as for this equation $x>0$
Now when $x <0$ equation becomes $x^2-x-6$
$(x+2)(x-3)=0$
Hence $x=-2$
So the roots are $2 and -2$
Thus sum of roots is zero and roots are real
So Option C

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

If $a, b, c$ are in A.P., then the roots of the equation $ax^{2}+2bx+c=0$ are

  1. real and distinct

  2. real and equal

  3. real

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

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

$2b=a+c$.
Hence the give equation transforms into
$ax^{2}+(a+c)x+c=0$
Hence
$D$
$=B^{2}-4AC$

$=(a+c)^{2}-4ac$

$=a^{2}+c^{2}+2ac-4ac$

$=a^{2}+c^{2}-2ac$

$=(a-c)^{2}$
Now 
$(a-c)^{2}\geq 0$
Hence 
$D\geq 0$.
Or 
$B^{2}-4AC\geq 0$.
Since the discriminant is greater than 0, hence the roots real.

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

If $a$ and $b$ are the roots of the quadratic equation $x^2-4x+3=0$, then $(1+a+a^2+a^3...)(1+b+b^2+b^3+....)$ equal to

  1. $\infty$
  2. $\dfrac{1}{4}$
  3. $\dfrac{-1}{6}$

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

${ x }^{ 2 }-4x+3=0\ (x-1)(x-3)=0\ x=1\quad or\quad 3\ \therefore a=3,\quad b=1\ a+b=4,\quad ab=3\ 1+a+{ a }^{ 2 }+{ a }^{ 3 }+........+\infty =\cfrac { 1 }{ 1-a } (sum\quad of\quad infinite\quad G.P.)\ \therefore (1+a+{ a }^{ 2 }+{ a }^{ 3 }+........+\infty )(1+b+{ b }^{ 2 }+{ b }^{ 3 }+.......+\infty )\ =(\cfrac { 1 }{ 1-a } )(\cfrac { 1 }{ 1-b } )\ =\cfrac { 1 }{ 1-(a+b)+ab } \ =\cfrac { 1 }{ 1-4+3 } =\cfrac { 1 }{ 0 } =\infty $

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

If $a,b,c,d$ are four consecutive terms of an increasing A.P., then the roots of the equation
$(x-a)(x-c)+2(x-b)(x-d)=0$ are

  1. $\text{real and distinct}$
  2. $\text {non-real complex}$
  3. $\text {real and equal}$
  4. $\text {integers}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$a,b,c,d$ are $4$ consecutive terms of A

Let $a=m-3n,b=m-n.c=m+n,d=m+3n$

$'2n'$ is common difference

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

$3{ x }^{ 2 }-\left( a+c+2b+2d \right) x+\left( ac+2bd \right) =0$

$=6m+2n$

$ac+2bd={ m }^{ 2 }-2mn-3{ n }^{ 2 }+2{ m }^{ 2 }+4mn-6{ n }^{ 2 }$

$=3{ m }^{ 2 }+2mn-9{ n }^{ 2 }$

$3{ x }^{ 2 }-\left( 6m+2n \right) x+\left( 3{ m }^{ 2 }+2mn-9{ n }^{ 2 } \right) =0$

$\triangle ={ \left( 6m+2n \right)  }^{ 2 }-4\left( 3 \right) \left( 3{ m }^{ 2 }+2mn-9{ n }^{ 2 } \right) $

$=4\left( 9{ m }^{ 2 }+6mn+{ n }^{ 2 }-9{ m }^{ 2 }-6mn+36{ n }^{ 2 } \right) $

$=4\left( 37 \right) { n }^{ 2 }$

$\triangle >0$

$\therefore $ Roots of $(x-a)(x-c)+2(x-b)(x-d)$ are real and distinct.
Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

If roots of equation $ x^2 - (2n+ 18) x - n-1 = 0 ( n \epsilon Z ) $ are rational, then number of possible value of $n $ is :

  1. $1$
  2. $2$
  3. $0$

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

For roots to be rational, the discriminant D = (2n+18)^2 - 4(1)(-n-1) must be a perfect square. D = 4n^2 + 72n + 324 + 4n + 4 = 4n^2 + 76n + 328. Setting 4n^2 + 76n + 328 = k^2, we find integer solutions for n.

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

 Choose the correct answer from the alternatives given.
If $\alpha \, and \, \beta$ are the roots of the equation $x^2$ - 7x + 12 = 0, then $\alpha^2 \, + \, \beta^2$ equals.

  1. 19

  2. 25

  3. 14

  4. 24
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
: Let $\alpha \, and \, \beta$ are the roots of the equation
$ax^2 + bx + c = 0$
We know that,
$\displaystyle \alpha \, + \, \beta \, = \, \frac{-b}{a} \, = \, \frac{- (-7)}{1} \, = \, 7$
$\displaystyle \alpha^2 \, + \, \beta^2 \, = \, (\alpha \, + \, \beta)^2 \, - \, 2\alpha \beta$
$\displaystyle \alpha^2 \, + \, \beta^2 \, = \, (7)^2 \, - \, 2 \, \times \, 12 \, = \, 49 \, - \, 24 \, = \, 25$
Multiple choice the nth roots of unity complex numbers maths

If $(2 + i \sqrt 3)$ is a root of the equation $x^2 + px + q = 0$, where p and q are real, then (p, q) equals to

  1. $(4, 7)$
  2. $(-4, -7)$
  3. $(-4, 7)$
  4. $(4, -7)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since $2 + i \sqrt 3$ is one root, then other root will be $2 - i \sqrt 3$.
$\therefore x^2 + px + q = 0$ is given equatiion
$\therefore$ Sum of roots $= 2 + i \sqrt 3 + 2 - i \sqrt 3 = p$
$\therefore p = - 4$
Product of roots $q = 4 + 3 = 7$

Multiple choice the nth roots of unity complex numbers maths

If $2 + i$ and $\sqrt {5} - 2i$ are the roots of the equation $(x^{2} + ax + b)(x^{2} + cx + d) = 0$, where $a, b, c, d$ are real constants, then product of all roots of the equation is

  1. $40$
  2. $9\sqrt {5}$
  3. $45$
  4. $35$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$2 - i$ and $\sqrt {5} + 2i$ are other roots.
So, Product is $(2 + i)(2 - i)(\sqrt {5} + 2i)(\sqrt {5} - 2i)$
$= 5\times 9 = 45$