Tag: maths

Questions Related to maths

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

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

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

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

Show answer
Correct Option: D
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
$a(x-2)^2+b(x-2)+c=0$
$\Rightarrow a(x^2-4x+4)+bx-2b+c=0$
$\Rightarrow ax^2 +(b-4a) x+(4a-2b+c)=0$
$\Rightarrow ax^{ 2 }+x(b-4a)+4a-2b+c=0$

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

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

  4. none of these

Show answer
Correct Option: A
Explanation:
$\alpha , \beta$ are the roots of the equation $x^2 - 3x + 1 = 0$
$\Rightarrow  \alpha^2-3\alpha+1=0$ ------(1)
Let $\displaystyle\dfrac{1}{\alpha-2}=y$
$\Rightarrow\displaystyle \alpha=2+\dfrac{1}{y}$
From (1), we get
$\displaystyle\left(2+\dfrac{1}{y}\right)^2-3\left(2+\dfrac{1}{y}\right)+1=0$
$\Rightarrow\displaystyle \dfrac{(2y+1)^2}{y^2}-\dfrac{3(2y+1)}{y}+1=0$
$\Rightarrow y^2-y-1=0$
$\therefore$ The equation with roots $\displaystyle \dfrac{1}{\alpha - 2} , \dfrac{1}{\beta - 2}$ is $x^2-x-1=0$
Hence, option A.

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 :

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

  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. $\displaystyle \left ( \frac{\beta}{1+ \beta} \right )$

Show answer
Correct Option: C,D
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}$

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

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

  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. $x^{2}-p^{2}x+p^{2}q = 0$

Show answer
Correct Option: D
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 $x^{2}-p^{2}x+qp^{2} = 0$

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

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

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

  2. $3{x^2} + 2x + 1 = 0$

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

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

Show answer
Correct Option: C
Explanation:

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

$\alpha+\beta=2$
$\alpha\beta=3$
$x^2-\left(\cfrac{\alpha-1}{\alpha+1}+\cfrac{\beta-1}{\beta+1}\right)x+\left(\cfrac{\alpha-1}{\alpha+1}\right)\left(\cfrac{\beta-1}{\beta+1}\right)=0$
$\Rightarrow x^2(\alpha+1)(\beta+1)-x((\alpha-1)(\beta+1)+(\beta-1)(\alpha+1))+(\alpha-1)(\beta-1)=0$
$\Rightarrow x^2(\alpha\beta+(\alpha+\beta)+1)-x(\alpha\beta+\alpha-\beta-1+\alpha\beta+\beta-\alpha-1)+(\alpha\beta-(\alpha+\beta)+1)=0$
$\Rightarrow x^2(3+2+1)-x(3-1)+(3-2+1)=0$
$\Rightarrow 6x^2-2x+2=0$
$\Rightarrow 3x^2-x+1=0$

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

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

  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

Show answer
Correct Option: A
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
$\displaystyle x^{2}-2x-\frac{\left ( 4\lambda ^{2}+11 \right )}{4}= 0$

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

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

  1. $3x^{2}-19x+3=0$

  2. $3x^{2}+19x-3=0$

  3. $3x^{2}-19x-3=0$

  4. $x^{2}+5x+3=0$

Show answer
Correct Option: A
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$=\cfrac { \alpha  }{ \beta  } \times \cfrac { \beta  }{ \alpha  } =1$
${ x }^{ 2 }-\cfrac { 19 }{ 3 } x+1=0$
$\therefore 3{ x }^{ 2 }-19x+3=0$

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

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

  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. $ax^{2}-bx+a=0$

Show answer
Correct Option: A
Explanation:
$\tan A$ & $\tan B$ are roots 
Sum $=(\tan A+\tan B)$
Product $=\tan A\tan B$
$\angle C=90^o$
$\angle A+\angle B=90^o$
$\tan A \tan B=1$
$\tan B=1/\tan A$
$x^2-(\tan A+\tan B)x+1=0$
$x^2\dfrac {2}{\sin 2A}x+1=0$
$\Rightarrow \ x^2-\dfrac {c^2}{ab}x+1=0 \Rightarrow \ x^2 (ab)-c^2 (x)+1=0$
$\tan A+\tan B=\tan A+\dfrac {1}{\tan A}$
$=\dfrac {2\tan ^2 A+1}{2\tan A}=\dfrac {2}{\sin 2A}$
$\sin 2A=\dfrac {2\tan A}{1+\tan^2 A}$