Tag: theory of equations

Questions Related to theory of equations

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

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

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

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

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

  4. $-x^2-x-1=0$

Show answer
Correct Option: B
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 $x^2+x+ 1 =0$

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-

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

  1. $p$ and $q$

  2. $1/p$ and $1/q$

  3. $-p$ and $-q$

  4. $p + q$ and $p - q$

Show answer
Correct Option: C
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 $-p,-q$

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 $\dfrac {1}{\beta}$?

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

  1. $\dfrac {1}{x^{2}} + \dfrac {1}{p} = 0$

  2. $px^{2} + 1 = 0$

  3. $px^{2} - 1 = 0$

  4. $\dfrac {1}{x^{2}} - \dfrac {1}{p} = 0$

Show answer
Correct Option: B
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
${ x }^{ 2 }-0+\cfrac { 1 }{ p } =0\ { px }^{ 2 }+1=0$

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 ?

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

  1. $ bc = a^2 $

  2. $ bc = 36 a^2 $

  3. $ bc = 72 a^2 $

  4. $ bc = 108 a^2 $

Show answer
Correct Option: D
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 $[-a(p+q)][apq]=-a(-18)a6=108a^2$

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

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

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

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

  3. $ax^2 +(b+c) x+bc=0$

  4. $ax^2-(b+c)x-bc=0$

Show answer
Correct Option: B
Explanation:

From the first equation, we can conclude that
$\alpha+\beta=-\cfrac{b}a$  ...(i)

$\alpha\beta=\cfrac{c}a$      ...(ii)

Therefore, the new equation will be

$x^2-(\alpha+\beta+\alpha\beta)x+(\alpha+\beta)(\alpha\beta)=0$

Substituting the values from (i) and (ii), we get

$x^2-\left(\cfrac{-b+c}{a}\right)x+\left(\cfrac{-bc}{a^2}\right)=0$

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


Hence, the answer is
$a^2x^2+a(b-c)x-bc=0$

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

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

  1. $a+2b$

  2. $a+b+c$

  3. $ab+bc+ca$

  4. $abc$

Show answer
Correct Option: B
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
$r=a+b+c$

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 :

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

  1. $2x^2+3x+18=0$

  2. $x^2+6x-9=0$

  3. $x^2+6x+9=0$

  4. $x^2-6x+9=0$

Show answer
Correct Option: C
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
$x^2-\left(-6\right)x+9=0$
$x^2+6x+9=0$

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

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

  1. $4{ x }^{ 2 }+49x-118=0$

  2. $4{ x }^{ 2 }-49x-118=0$

  3. $4{ x }^{ 2 }-49x+118=0$

  4. $4{ x }^{ 2 }+49x+118=0$

Show answer
Correct Option: C
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
$x^{2}-\dfrac{49x}{4}+\dfrac{59}{2}=0$
$4x^{2}-49x+118=0$

If $\alpha, \beta$ are the roots of $x^2 + px+1=0$ and $\gamma, \delta $ are the roots of $x^2+qx+1=0$, then $(\alpha - \gamma) (\beta - \gamma)(\alpha - \delta) (\beta + \delta)=$

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

  1. $2q^2$

  2. $2p^2$

  3. $p^2-q^2$

  4. $q^2 - p^2$

Show answer
Correct Option: D
Explanation:

$\alpha, \beta$ are the roots of $x^2 + px+1=0$
$\Rightarrow \alpha+\beta = -p,  \alpha \beta =1$

$\gamma, \delta$ are the roots of $x^2+qx+1=0$
$\Rightarrow \gamma \delta =1, \gamma^2+q\gamma +1=0$,
$\delta^2 +q\delta +1=0$

$(\alpha - \gamma)(\beta -\gamma)(\alpha + \delta)(\beta + \delta)$
$=[\alpha \beta - \gamma (\alpha + \beta)+\gamma^2][\alpha \beta + \delta (\alpha + \beta) + \delta^2]$
$=(1+p \gamma + \gamma^2)(1-p\delta + \delta^2)$
$=(p \gamma - q \gamma)(-p \delta - q\delta)$
$=-\gamma \delta (p-q)(p+q)$
$=-(p^2-q^2) = q^2 - p^2$

Hence, option D.

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.

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

  1. $x^2 - 6x - 5 = 0$

  2. $x^2 - 5x - 6 = 0$

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

  4. None of the above

Show answer
Correct Option: A
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 $x^2 - 6x - 5 = 0$