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

If $\alpha, \beta $ are the roots of $ax^2+bx+c=0$ then the equation whose roots are $2+\alpha , 2+\beta$ is:

  1. $ax^2+x(4a-b) + 4a-2b+c=0$
  2. $ax^2+x(4a-b) + 4a+2b+c=0$
  3. $ax^2+x(b-4a) = 4a+2b+c=0$
  4. $ax^2+x(b-4a) + 4a-2b+c=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\alpha, \beta$ are the roots of $\Rightarrow { ax }^{ 2 }+bx+c=0$
Then, $a(\alpha)^2 + b(\alpha)+c =0$
Now, $\alpha +2 = x $
Hence, $\alpha = x -2$
Thus, replace $\alpha$ by $x-2$ in the given equation,
Required equation is
$a(x-2)^2+b(x-2)+c=0$
$\Rightarrow a(x^2-4x+4)+bx-2b+c=0$
$\Rightarrow ax^2 +(b-4a) x+(4a-2b+c)=0$
$\Rightarrow ax^{ 2 }+x(b-4a)+4a-2b+c=0$

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

If $\alpha, \beta$ are roots of $ax^2+bx+c=0$, then one root of the equation $ax^2-bx(x-1) + c(x-1)^2=0$ is :

  1. $\displaystyle \left ( \frac{\alpha}{1- \alpha} \right )$
  2. $\displaystyle \left ( \frac{1-\beta}{\beta} \right )$
  3. $\displaystyle \left ( \frac{\alpha}{1+ \alpha} \right )$
  4. $\displaystyle \left ( \frac{\beta}{1+ \beta} \right )$
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

We have, $ax^2-bx^2+bx+cx^2-2cx+c=0$

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

Sum of the roots (S)
$\displaystyle \frac{b-2c}{a-b+c} = \frac{\left ( -\frac{b}{a} + \frac{2c}{a} \right )}{\left ( 1- \frac{b}{a} + \frac{c}{a}\right )}$

$\displaystyle S = \frac{\alpha+\beta+1\alpha \beta}{2+ \alpha + \beta+ \alpha \beta}=\frac{\alpha}{\alpha+1}+ \frac{\beta}{\beta+1}$

Product of the roots (P) $\displaystyle =\frac{c}{a-b+c}$

$\Rightarrow \displaystyle P= \frac{\left ( \frac{c}{a} \right )}{\left ( 1- \frac{b}{c}+\frac{c}{a}\right )}$

$\displaystyle=\frac{\alpha \beta}{1+\alpha+\beta+\alpha \beta} = \frac{\alpha}{(\alpha+1)} \cdot \frac{\beta}{(\beta+1)}$

Thus the roots are $ \displaystyle \frac{\alpha}{\alpha+1} and \frac{\beta}{\beta+1}$

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

If $\alpha $ and $\beta$ be the roots of the equation $x^{2}+px+q = 0$, then the equation whose roots are $\alpha^{2}+\alpha\beta$ and $\beta^{2}+\alpha\beta$ is

  1. $x^{2}+p^{2}x+p^{2}q = 0$
  2. $x^{2}-q^{2}x+p^{2}q = 0$
  3. $x^{2}+q^{2}x+p^{2}q = 0$
  4. $x^{2}-p^{2}x+p^{2}q = 0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since $\alpha$ and $\beta$ are roots of the equation
$x^{2}+px+q = 0$, therefore
$\alpha + \beta = -p$      ...(i)
and $\alpha\beta = q$      ...(ii)
Sum of the roots $= \alpha^{2}+\alpha\beta+\beta^{2}+\alpha\beta$
$= (\alpha+\beta)^{2} = p^{2}$
Product of the roots $=(\alpha^{2}+\alpha\beta)(\beta^{2}+\alpha\beta)$
$= \alpha\beta (\alpha+\beta)^{2} = qp^{2}$
Required equation will be
$x^{2}$-(Sum  of  the  roots)$x$ + Product  of  the  roots = $0$
or $x^{2}-p^{2}x+qp^{2} = 0$

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

Find a quadratic equation whose roots $\displaystyle \alpha$ and $ \displaystyle \beta $ are connected by the relation:
$\displaystyle \alpha +\beta = 2$ and $\displaystyle \frac{1-\alpha }{1+\beta }+\frac{1-\beta }{1+\alpha }= 2\left ( \frac{4\lambda ^{2}+15}{4\lambda ^{2}-1} \right )$

  1. $\displaystyle x^{2}-2x-\frac{\left ( 4\lambda ^{2}+11 \right )}{4}= 0$
  2. $\displaystyle x^{2}+2x-\frac{\left ( 4\lambda ^{2}-11 \right )}{4}= 0$
  3. $\displaystyle x^{2}-2x+\frac{\left ( -2\lambda ^{2}+11 \right )}{4}= 0$

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

$\displaystyle \alpha +\beta = 2$ and let $\displaystyle \alpha \beta = p$
$\displaystyle \therefore $ Equation is $\displaystyle x^{2}-2x+p= 0$ ...(1)
We have to find the value of p.
Now $\displaystyle

\frac{1-\alpha }{1+\beta }+\frac{1-\beta }{1+\alpha }= \frac{\left (

1-\alpha ^{2} \right )+\left ( 1-\beta ^{2} \right )}{1+\left ( \alpha

+\beta  \right )+p}$
or $\displaystyle \frac{2-\left ( \alpha

^{2}+\beta ^{2} \right )}{1+2+p}= \frac{2-\left { \left ( \alpha +\beta

 \right )^{2}-2\alpha \beta  \right }}{3+p}$
or $\displaystyle

\frac{2-4+2p}{3+p}:or:\frac{2\left ( p-1 \right )}{p+3}= 2\left (

\frac{4\lambda ^{2}+15}{4\lambda ^{2}-1} \right )$
or $\displaystyle \frac{p-1}{p+3}= \frac{4\lambda ^{2}+15}{4\lambda ^{2}-1}$
or $\displaystyle

p\left [ \left ( 4\lambda ^{2}-1 \right )-\left ( 4\lambda ^{2}+15

\right ) \right ]= 3\left ( 4\lambda ^{2}+15 \right )+\left ( 4\lambda

^{2}-1 \right )$
or $\displaystyle -16p= 16\lambda ^{2}+44= 4\left ( 4\lambda ^{2}+11 \right )$
$\displaystyle \therefore p= -\frac{\left ( 4\lambda ^{2}+11 \right )}{4}$
Putting for $p$ in (1) we get the required equation as
$\displaystyle x^{2}-2x-\frac{\left ( 4\lambda ^{2}+11 \right )}{4}= 0$

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

If $\alpha \neq \beta, \alpha^{2}=5\alpha -3$, and $\beta^{2}=5\beta-3$, then the equation having $\alpha/\beta$ and $\beta/\alpha$ as its roots is

  1. $3x^{2}-19x+3=0$
  2. $3x^{2}+19x-3=0$
  3. $3x^{2}-19x-3=0$
  4. $x^{2}+5x+3=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
${ \alpha  }^{ 2 }=5\alpha -3\quad { \beta  }^{ 2 }=5\beta -3$
Equation is ${ x }^{ 2 }-5x+3=0$
$\alpha +\beta =5\quad \alpha \beta =3$
If $\cfrac { \alpha  }{ \beta  } ,\cfrac { \beta  }{ \alpha  } $ are roots
Sum of roots$=\cfrac { \alpha  }{ \beta  } +\cfrac { \beta  }{ \alpha  } =\cfrac { { \alpha  }^{ 2 }+{ \beta  }^{ 2 } }{ \alpha \beta  } =\cfrac { { \left( \alpha +\beta  \right)  }^{ 2 }-2\alpha \beta  }{ \alpha \beta  } $
$=\cfrac { 25-2\left( 3 \right)  }{ 3 } =\cfrac { 19 }{ 3 } $
Products of roots$=\cfrac { \alpha  }{ \beta  } \times \cfrac { \beta  }{ \alpha  } =1$
${ x }^{ 2 }-\cfrac { 19 }{ 3 } x+1=0$
$\therefore 3{ x }^{ 2 }-19x+3=0$
Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

In a $\triangle ABC, C=90^{o}$. Then $\tan A$ and $\tan B$ are the roots of the equation

  1. $abx^{2}-c^{2}x+1=0$
  2. $abx^{2}-(a^{2}+b^{2})x+ab=0$
  3. $c^{2}x^{2}-abx+c^{2}=0$
  4. $ax^{2}-bx+a=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\tan A$ & $\tan B$ are roots 
Sum $=(\tan A+\tan B)$
Product $=\tan A\tan B$
$\angle C=90^o$
$\angle A+\angle B=90^o$
$\tan A \tan B=1$
$\tan B=1/\tan A$
$x^2-(\tan A+\tan B)x+1=0$
$x^2\dfrac {2}{\sin 2A}x+1=0$
$\Rightarrow \ x^2-\dfrac {c^2}{ab}x+1=0 \Rightarrow \ x^2 (ab)-c^2 (x)+1=0$
$\tan A+\tan B=\tan A+\dfrac {1}{\tan A}$
$=\dfrac {2\tan ^2 A+1}{2\tan A}=\dfrac {2}{\sin 2A}$
$\sin 2A=\dfrac {2\tan A}{1+\tan^2 A}$


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

If $\displaystyle \alpha $ are $\displaystyle \beta $ are the roots of $\displaystyle x^{2}+x+1=0$  then find the equation whose roots $\displaystyle \alpha ^{2}$ and $\displaystyle \beta ^{2}$

  1. $\displaystyle x^{2}+x+1=0$
  2. $\displaystyle x^{2}+2x+1=0$
  3. $\displaystyle x^{2}+x+2=0$
  4. $\displaystyle x^{2}+2x+2=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the given equation, sum of roots $ = \alpha + \beta  = -\dfrac {1}{1} = -1 $

Product of roots $ = \alpha \times \beta =  \dfrac {1}{1} =1 $

Now, $ {\alpha}^{2} + \beta ^{2} = (\alpha + \beta )^{2} - 2(\alpha \times \beta)= (-1)^{2} - 2(1) = 1-2 = -1 $

And $ {\alpha}^{2} \times \beta ^{2} = (\alpha \times \beta )^{2} = 1 $

Equation whose roots are $ {\alpha}^{2} $ and $ \beta ^{2} $ is $ x^{2} -(Sum \ of \ roots)x +  Product \ of \ roots  = 0 $
$ => x^{2} -({\alpha}^{2} + \beta ^{2})x +  {\alpha}^{2} \times \beta ^{2}  = 0 $
$ => x^{2} -(-1)x+ 1  = 0 $
$ => x^{2} +x+ 1  = 0 $

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

Two students Ragini and Gourav were asked to solve a quadratic equation $\displaystyle ax^{2}+bx+c=0,a\neq 0$ Ragini made some mistake in writing b and found the roots as 3 and $\displaystyle -\frac{1}{2}$ Gourav too made mistake in writing c and found the roots -1 and $\displaystyle -\frac{1}{4}$ The correct roots of the given equation should be

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

Given: Ragini found roots as $3, -\dfrac 12$ when copied the wrong coefficient of $x$ and Gourav found roots as $-1, -\dfrac 14$ when copied wrong constant term.

To find the correct roots of the given equation
Sol: Viete's formula for the roots $x _1$ and $x _2$ of equation $ax^2+bx+c=0$:  $x _1+x _2=−\dfrac ba$ and $ x _1\times x _2=\dfrac ca$
According to Ragini, she copied the constant term and the coefficient of $ x^2$ correctly. Hence $3\times -\dfrac 12=-\dfrac 32=\dfrac ca$
And according to Gourav, he copied coefficient of $x$ and $x^2$ correctly. Hence $-1+\left(-\dfrac 14\right)=-\dfrac 54=-\dfrac ba$
Hence the equation becomes,
$x^2+\dfrac 54x-\dfrac 32=0\implies 4x^2+5x-6=0$
$\implies 4x^2+8x-3x-6=0\\implies 4x(x+2)-3(x+2)=0\\implies (4x-3)(x+2)=0\\implies x _1=-2, x _2=\dfrac 34$
are the correct roots of the given equation.

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

Rohan and Sohan were attempting to solve the quadratic equation  $\displaystyle x^{2}-ax+b=0$. Rohan copied the coefficient of x wrongly and obtained the roots as 4 and 12 . Sohan copied the constant term wrongly and obtained the roots as -19 and 3. Find the correct roots

  1. -8, -10

  2. -8, -6

  3. -4, -12

  4. 4, 12
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

With the roots $ 4, 12 $, the equation was $ (x-4))(x-12) ={x}^{2} -4x  -12x + 48 = {x}^{2} -16x + 48 $

As Rohan made a mistake in noting the coffecient of $ x $ , in the original equation, coefficient of $ {x}^{2} = 1 $ and constant $ = 48 $

Now, with the roots $ -19, 3 $, the equation was $ (x-(-19))(x-3) = (x+19)(x-3) = {x}^{2} + 19x -3x -57 = {x}^{2} + 16x -57 $

As Sohan made a mistake in noting just the constant term in the original  equation, coefficient of $ {x}^{2} = 1 $ and of $ x = 16 $

So, we get the original equation as $ {x}^{2} + 16x + 48 = 0 $
Solving it, we get $ {x}^{2} + 4x +12x + 48 = 0 $
$ => x(x+4) + 12(x+4) = 0 $
$ => (x+4)(x+12) = 0 $
$ => x = -4, -12 $

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

If the equation formed by decreasing each root of $ax^{2}+bx+c=0$ by $1$ is $2x^{2}+8x+2=0$, then

  1. $\mathrm{a}=-\mathrm{b}$
  2. $\mathrm{b}=-\mathrm{c}$
  3. $\mathrm{c}=-\mathrm{a}$
  4. $\mathrm{b}=\mathrm{a}+\mathrm{c}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since the equation $2{ x }^{ 2 }+8x+2=0$ has roots which are 1 less than those of the equation $a{ x }^{ 2 }+bx+c=0$ then if we replace $ x $ by $ x+1 $ in latter we'll get the former.
$\Rightarrow a(x+1)^{ 2 }+b(x+1)+c=0$
$\Rightarrow  a{ x }^{ 2 }+(2a+b)x+a+b+c=0$
comparing this equation with that of $2{ x }^{ 2 }+8x+2=0$
we get option (b) as the correct answer

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

Umesh and Varun are solving an equation of the form $\displaystyle x^{2}+bx+c=0$. In doing so Umesh commits a mistake in noting down the constant term and finds the roots as $-3$ and $-12$. And Varun commits a mistake in noting down the coefficient of $x$ and find the roots as $-27$ and $-2$. If so find the original equation

  1. $\displaystyle x^{2}-15x+36=0$
  2. $\displaystyle x^{2}+15x+36=0$
  3. $\displaystyle x^{2}-15x+54=0$
  4. $\displaystyle x^{2}+15x+54=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

With the roots $ -3, -12 $, the equation was $ (x-(-3))(x-(-12) = (x+3)(x+6) = {x}^{2} + 3x + 12x + 36 = {x}^{2} + 15x + 36 $

As Umesh made a mistake in noting just the constant term, in the original equation, coefficient of $ {x}^{2} = 1 $ and of $ x = 15 $

Now, with the roots $ -27, -2 $, the equation was $ (x-(-27))(x-(-2)) = (x+27)(x+2) = {x}^{2} + 27x + 2x + 54 = {x}^{2} + 29x + 54 $

As Varun made a mistake in noting the coffecient of $ x $ in the original  equation, coefficient of $ {x}^{2} = 1 $ and constant $ = 54 $

So, we get the original equation as $ {x}^{2} + 15x + 54 = 0 $

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

If a,b,c and d are the real roots of the equation : $x^{4}+p _{1}x^{1}+p _{2}x^{2}+p _{3}x+p _{4}=0$ and $(1+a^{2})(1+b^{2})(1+c^{2})(1+d^{2})=k(1-p _{2}+p _{4})^{2}+(p _{3}-p _{1})^{2}$ then the value f k is:

  1. -1

  2. 1

  3. 2

  4. -2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the relationship between roots and coefficients, and the identity that (1+a²)(1+b²)(1+c²)(1+d²) can be expressed in terms of the polynomial's coefficients, we can derive that k = 1. This involves substituting the roots into the given expression and using Vieta's formulas to relate it to the coefficients p₁, p₂, p₃, p₄.

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

If $\alpha $ and $\beta $ are the roots of ${ x }^{ 2 }+px+q=0$ and ${ \alpha  }^{ 4 } , { \beta  }^{ 4 }$ are the roots of ${ x }^{ 2 }-rx+s=0$, then the equation ${ x }^{ 2 }-4qx+2{ q }^{ 2 }-r=0$ has always two real roots.

  1. True

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

If alpha, beta are roots of x^2 + px + q = 0, then alpha^4, beta^4 are roots of x^2 - rx + s = 0. The discriminant of the final equation x^2 - 4qx + 2q^2 - r = 0 is D = (4q)^2 - 4(2q^2 - r) = 16q^2 - 8q^2 + 4r = 8q^2 + 4r. Since r = alpha^4 + beta^4, which is always positive, D > 0, so the roots are real.

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

Suppose $a$ and $b$ are real no. such that the roots of the cubic equation $ax^{3}-x^{2}+bx+1=0$ are all positive real no. then
$0 < 3ab \le 1$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The given equation is $ax^3-x^2+bx+1=0$
Let $\alpha,\,\beta,\,\gamma$ be the roots of the given equation.
We have
$\alpha+\beta+\gamma=\dfrac{1}{a}$
$\alpha\beta+\beta\gamma+\gamma\alpha=\dfrac{b}{a}$
$\alpha\beta\gamma=\dfrac{1}{a}$
It follows that $a,b$ are positive. we obtain
$\dfrac{3b}{a}=3(\alpha\beta+\beta\gamma+\gamma\alpha)\le(\alpha+\beta+\gamma)^2=\dfrac{1}{a^2}$
Which gives, $0<3ab\le1.$
Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

If the sum of two roots of the equation $\displaystyle x^{3}+ax^{2}+bx+c= 0 $ is zero, then value of $ab$ equals

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

Given equation is $x^{3}+ax^{2}+bx+c= 0$

Let the roots be $\alpha, -\alpha, \beta$

Then $\alpha-\alpha+\beta=-a$

$\Rightarrow \beta=-a$       ....(1)

Also, $-{\alpha}^{2}+{\alpha}\beta-\alpha\beta=b$

$\Rightarrow -{\alpha}^{2}=b$       .....(2)

Also, $-{\alpha}^{2} \beta=-c$  ....(by (1)and (2))

$\Rightarrow -ab=-c$

$\displaystyle \Rightarrow ab= c $