Quantitative Aptitude

Algebra

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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 one root of the quadratic equation $ax^{2}\, +\, bx\, +\, c\, =\, 0$ is the square of the other, then $b^{3}\, +\, a^{2}c\, +\, ac^{2}\, =\, 3abc$
Say yes or no.

  1. Yes

  2. No

  3. Ambiguous

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

Let one root of $ax^2 + bx + c =0$ be $\alpha$ and other be $\alpha^2$
then, $\alpha + \alpha^2 = \frac{-b}{a}$
$\alpha^3 = \frac{c}{a}$
or $\alpha = (\frac{c}{a})^{({\frac{1}{3}})}$
Now put the value of $\alpha$ in $\alpha + \alpha^2 = \frac{-b}{a}$
$\frac{c}{a}^{\frac{1}{3}} + \frac{c}{a}^{\frac{2}{3}} = \frac{-b}{a}$
Cubing both sides:

$(\frac{c}{a})^{({\frac{3}{3}})} + (\frac{c}{a})^{({\frac{6}{3}})} + 3 {(\frac{c}{a})^{({\frac{1}{3}})}}\times{(\frac{c}{a})^{({\frac{2}{3}})}}((\frac{c}{a})^{({\frac{1}{3}})} + (\frac{c}{a})^{({\frac{2}{3}})}) = (\frac{-b}{a})^3$

$\frac{c}{a} + \frac{c^2}{a^2} + 3\frac{c}{a}(\frac{-b}{a}) = \frac{-b^3}{a^3}$

$a^2c + ac^2 - 3abc = - b^3 $
$b^3 + a^2c + ac^2 = 3abc$

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

If the roots of the equation $2x^2 - 3x + 5 = 0$ are reciprocals of the roots of the equation $ax^2 + bx + 2 = 0$, then

  1. $a = 2, b = 3$
  2. $a = 2, b = -3$
  3. $a = 5, b = -3$
  4. $a = 5, b = 3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $\alpha ,\beta $ are roots of $2{ x }^{ 2 }-3x+5=0$ 
Then to get equation whose roots are $\displaystyle \dfrac { 1 }{ \alpha  } ,\dfrac { 1 }{ \beta  } $ 
Replace $\displaystyle x\rightarrow \frac { 1 }{ x } \ $
We get $5{ x }^{ 2 }-3x+2=0$.

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

If each root of the equation ${x}^{2}+11{x}+13=0$ is diminished by $4$, then the resulting equation is

  1. ${x}^{2}+3{x}-15=0$
  2. ${x}^{2}+3{x}+73=0$
  3. ${x}^{2}+19{x}+73=0$
  4. ${x}^{2}-3{x}-4=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$ The\quad roots\quad of\quad the\quad equation\quad { x }^{ 2 }+11x+13=0\quad is\quad diminished\quad by\quad 4\quad i.e.\ the\quad variable\quad becomes\quad (x+2).\ \therefore \quad
The\quad new\quad equation\quad is\quad \ (x+4)^{ 2 }+11(x+4)+13=0\ \Rightarrow { x }^{ 2 }+8x+16+11x+44+13=0\ \Rightarrow { x }^{ 2 }+19x+73=0\ Ans-\quad Option\quad C .$

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, \beta $ are the roots of $\displaystyle x^{2}+3x+3=0$  then find the quadratic equation whose roots are $\displaystyle (\alpha +\beta )$ and $\displaystyle \alpha \beta $

  1. $\displaystyle x^{2}=1$
  2. $\displaystyle x^{2}=4$
  3. $\displaystyle x^{2}=9$

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

For the equation, sum of roots $ = \alpha  + \beta = -\dfrac {3}{1} = -3 $
Product of roots $ = \alpha  \times \beta = \dfrac {3}{1} = 3 $

So, the eqn with roots $ = \alpha  + \beta$ and $ \alpha  \times \beta $ is $ (x - (-3))(x-3) = 0 $
$ => x^{2} -9 = 0 $

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

If $a, b, g$  are the roots of the equation $(x - 2$ ) $\displaystyle \left ( x^{2}+6x-11 \right )=0$ therefore $(a + b + g)$  equals

  1. $-4$
  2. $\dfrac{23}{6}$
  3. $13$
  4. $-8$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, $(x-2)(x^{2}+6x-11)=0$
$x^{3}+6x^{2}-11x-2x^{2}-12x-22=0$
$x^{3}+4x^{2}-23x-22=0$
Then $a=1  ,b=4  g=-22$
Sum of roots $(a+b+g) =\displaystyle \frac{-b}{a}=\frac{-4}{1}=-4$

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

The roots of equation $\displaystyle x^{2}+px+q=0$ are $1 $ and $2$ . The roots of the equation $\displaystyle qx^{2}-px+1=0$ must be

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

The equation can be written as $x^{ 2 }+px+q=0$

The roots are $1$ and $2$
Sum of roots $= 3= -p$
Product of roots $= 2= q$
The second equation is $2x^{ 2 }+3x+1=0$
$2x^{ 2 }+2x+x+1=0$
$ \Longrightarrow 2x(x+1)+1(x+2)=0$
$\Longrightarrow (x+1)(2x+1)=0$
$ \Longrightarrow x=-1 $ or $-1/2$

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

The equation whose roots are twice the roots of $x^2 -3x +3=0$ is

  1. $x^2-6x+12=0$
  2. $x^2-3x+6=0$
  3. $2x^2-3x+3=0$
  4. $4x^2-6x+3=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$x^2 -3x = 3 = 0$ ........ (1)
Here, $a +\beta =3$ and $a\beta =3$. Therefore,
$2(a +\beta ) =6$
$2\times 2a\beta = 4a\beta =4 \times 3= 12$
The equation whose roots are double of (1),
$x^2 -$(sum of the roots)x + product of the roots $=0$
will be $x^2 -6x + 12 =0$

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

The equation whose roots are the squares of the roots of equation $x^2 -x +1= 0$ is

  1. $x^2-x+1=0$
  2. $x^2+x+1=0$
  3. $x^2-x-1=0$
  4. $-x^2-x-1=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The given equation is $x^2 -x + 1 = 0$ ....... (1)
Here, $a +\beta = 1$ and $a\beta = 1$. Therefore,
$a^2+\beta^2 = (a + \beta)^2 -2a\beta = 1-2= -1$
and $a^2\beta^2 = (a\beta)^2 = 1^2= 0$
Therefore, the equation whose roots are square of 1 is $x^2$ -(sum of the roots)x +product $=0$
or $x^2-(-1)x+ 1 =0$
or $x^2+x+ 1 =0$

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

If $m$ and $n$ are the roots of the equation $(x + p)(x + q) - k = 0$, then the roots of the equation $(x - m)(x - n) + k = 0$ are-

  1. $p$ and $q$
  2. $1/p$ and $1/q$
  3. $-p$ and $-q$
  4. $p + q$ and $p - q$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$(x+p)(x+q)-k=0\ \Longrightarrow { x }^{ 2 }+(p+q)x+pq-k=0$

$m$ and $n$ are the roots of this equation
So, we have
Sum of roots $= -(p+q)=m+n$
Product of the roots $=pq-k= mn$
$\Rightarrow pq=mn+k$
Consider, $(x-m)(x-n)+k=0$ 
$\Rightarrow { x }^{ 2 }-(m+n)x+mn+k=0$
Sum of roots is $ m+n$
But $m+n= (-p)+(-q)$
Product of the roots $=mn+k$
But $mn+k= pq= (-p)(-q)$
Hence, the roots of the new equation are $-p,-q$

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 $\beta$ are the roots of $x^{2} + p = 0$ where p is a prime, which equation has the roots $\dfrac {1}{\alpha}$ and $\dfrac {1}{\beta}$?

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

${ x }^{ 2 }+p=0$

roots are $\alpha & \beta $
sum = $\alpha +\beta =0$
product=$\alpha \beta =p$
New roots are $\cfrac { 1 }{ \alpha  } & \cfrac { 1 }{ \beta  } $
sum = $\cfrac { 1 }{ \alpha  } +\cfrac { 1 }{ \beta  } =\cfrac { \alpha +\beta  }{ \alpha \beta  } =0\ $
product = $\cfrac { 1 }{ \alpha \beta  } =\cfrac { 1 }{ p } $
equation 
${ x }^{ 2 }$-(sum of roots)x+product of roots = 0
${ x }^{ 2 }-0+\cfrac { 1 }{ p } =0\ { px }^{ 2 }+1=0$

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

The equation formed by multiplying each root of $ax^2  + bx + c = 0$ by 2 is $ x^2 + 36x + 24 = 0$.Which one of the following is correct ?

  1. $ bc = a^2 $
  2. $ bc = 36 a^2 $
  3. $ bc = 72 a^2 $
  4. $ bc = 108 a^2 $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

let $p,q$ be roots of equation $ax^2+bx+c=0$


So $p+q=\left(-\dfrac{b}{a}\right)$ and $pq=c/a$


$\Rightarrow b=-a(p+q),c=apq$

New equation is $x^2+36x+24=0$ and roots are $2p,2q$

So $2p+2q=-36$

$\Rightarrow p+q=-18$

$2p\times 2q=24$

$\Rightarrow pq=6$

Then, value of $bc$ is $[-a(p+q)][apq]=-a(-18)a6=108a^2$

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 $\beta$ are the roots of the equation $ax^2+bx+c=0$ and if $px^2+qx+r=0$ has roots $\displaystyle \frac{1-\alpha}{\alpha}$ and $\displaystyle \frac{1-\beta}{\beta}$, then $r$ is

  1. $a+2b$
  2. $a+b+c$
  3. $ab+bc+ca$
  4. $abc$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
The equation with roots $\cfrac1{\alpha}$ and $\cfrac1{\beta}$
$=a\left(\cfrac1x\right)^2+b\left(\cfrac1x\right)+c$
$=cx^2+bx+a=0$ ..(1)
Now $\cfrac{1-\alpha}{\alpha}=\cfrac{1}{\alpha}-1$
Similarly $\cfrac{1-\beta}{\beta}=\cfrac{1}{\beta}-1$
Therefore the quadratic equation containing these roots is
$c\left(x+1\right)^2+b\left(x+1\right)+a$
$=cx^2+\left(b+2c\right)x+a+b+c = 0$
By comparing coefficients we get with $px^2+qx+r=0$ we get
$r=a+b+c$
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 the equation $9x^2+6x+1=0$, then the equation with the roots $\cfrac{1}{\alpha}, \cfrac{1}{\beta}$ is :

  1. $2x^2+3x+18=0$
  2. $x^2+6x-9=0$
  3. $x^2+6x+9=0$
  4. $x^2-6x+9=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$\left(x-\cfrac1{\alpha}\right)\left(x-\cfrac1{\beta}\right)=0$
$x^2-\left(\cfrac{1}{\alpha}+\cfrac{1}{\beta}\right)x+\left(\cfrac{1}{\alpha\beta}\right)=0$
$x^2-\left(\cfrac{\alpha+\beta}{\alpha\beta}\right)x+\left(\cfrac{1}{\alpha\beta}\right)=0$
From the given equation we know
$\alpha+\beta=-\cfrac69$
$\alpha\beta=\cfrac19$
By substituting we get
$x^2-\left(-6\right)x+9=0$
$x^2+6x+9=0$
Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $\alpha$ and $\beta$ are roots of $2{ x }^{ 2 }-3x-6=0$, then the equation whose roots are ${ \alpha  }^{ 2 }+2$ and ${ \beta  }^{ 2 }+2$ will be

  1. $4{ x }^{ 2 }+49x-118=0$
  2. $4{ x }^{ 2 }-49x-118=0$
  3. $4{ x }^{ 2 }-49x+118=0$
  4. $4{ x }^{ 2 }+49x+118=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$2x^{2}-3x-6=0$

$\alpha+\beta=\dfrac{3}{2}$
$\alpha\beta=-3$
Now roots are $\alpha^{2}+2$   and   $\beta^{2}+2$
$sum=\alpha^{2}+2+\beta^{2}+2$
$=(\alpha+\beta)^{2}-2\alpha\beta+4$
$=\dfrac{9}{4}+6+4$
$=\dfrac{49}{4}$
$Product=(\alpha^{2}+2)(\beta^{2}+2)$
$=\alpha^{2}\beta^{2}+2(\alpha^{2}+\beta^{2})+4$
$=9+2\times\dfrac{33}{4}+4=13+\dfrac{33}{2}=\dfrac{59}{2}$
Equation
$x^{2}-\dfrac{49x}{4}+\dfrac{59}{2}=0$
$4x^{2}-49x+118=0$

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

Find the equation whose sum of roots and product of roots are the product and sum of roots of $x^2 + 5x + 6 = 0$ respectively.

  1. $x^2 - 6x - 5 = 0$
  2. $x^2 - 5x - 6 = 0$
  3. $x^2 + 11x - 1 = 0$

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

In the given equation
sum of roots $= -5$ and product of roots $= 6$
The standard form of a quadratic equation is: $x^2 - (S)x + P = 0$, where S and P are sum and product of roots.
So according to question
$x^2 - (6)x + (-5) = 0$
The required equation will be $x^2 - 6x - 5 = 0$