Tag: maths

Questions Related to maths

If $\alpha$ and $\beta$ are the roots of the equation $4x^{2}\, -\, 5x\, +\, 2\, =\, 0$, find the equation whose roots are
$\displaystyle \frac{\alpha}{\beta}$ and $\displaystyle \frac{\beta}{\alpha}$.

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

  1. $8x^{2}\, +\, 9x\, +\, 8\, =\, 0$

  2. $8x^{2}\, -\, 9x\, +\, 8\, =\, 0$

  3. $8x^{2}\, -\, 9x\, -\, 8\, =\, 0$

  4. $x^{2}\, -\, 9x\, +\, 8\, =\, 0$

Show answer
Correct Option: B
Explanation:

$4x^2 - 5x + 2 = 0 $


If $\alpha$ and $\beta$ are the roots of this equation, 

then , sum of roots: $\alpha + \beta$ = $\dfrac{5}{4}$

Product of roots: $\alpha. \beta = \dfrac{2}{4}$

The equation which has roots as : $\dfrac{\alpha}{\beta}$ and $\dfrac{\beta}{\alpha}$

Sum of roots: $\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha}$ 

= $\dfrac{\alpha^2 + \beta^2}{\alpha\beta}$

= $\dfrac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta}$

= $\dfrac{ \left (\dfrac{5}{4} \right )^2 - 2\dfrac{2}{4}}{\dfrac{2}{4}}$

= $\dfrac{9}{8}$

Product of roots: $ \left (\dfrac{\alpha}{\beta} \right ) \left (\dfrac{\beta}{\alpha} \right )= 1$

Thus new equation is :$x^2 -Sx + P = 0$

$x^2 - \dfrac{9}{8} + 1= 0$

$8x^2 - 9x + 8 = 0$

Let $\alpha$ and $\beta$ be the roots of the equation ${ x }^{ 2 }+x+1=0$. The equation whose roots are ${ \alpha  }^{ 19 },{ \beta  }^{ 7 }$ 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: D
Explanation:

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


$\Rightarrow \left( x-\omega  \right) \left( x-{ \omega  }^{ 2 } \right) =0$

$\Rightarrow x=\omega ,{ \omega  }^{ 2 }$

$\therefore \alpha =\omega ,\beta ={ \omega  }^{ 2 }$   $(\because \omega ,{ \omega  }^{ 2 }$ are cube roots of unity $)$

Hence, ${ \alpha  }^{ 3 }=\omega ^3 =1$
             ${ \beta  }^{ 3 }=[{\omega ^3}]^2 = 1$
             $\alpha \beta =\omega^3=1$

$\therefore { \alpha  }^{ 19 }={ \left( { \alpha  }^{ 3 } \right)  }^{ 6 }\alpha ={ 1 }^{ 6 }\alpha =\alpha =\omega $ and ${ \beta  }^{ 7 }={ \beta  }^{ 6 }.\beta ={ 1 }^{ 2 }.\beta =\beta ={ \omega  }^{ 2 }$

$\\ \Rightarrow { \alpha  }^{ 19 }+{ \beta  }^{ 7 }=\omega +{ \omega  }^{ 2 }=-1$ 
$\Rightarrow { \alpha  }^{ 19 }{ \beta  }^{ 7 }=\omega .{ \omega  }^{ 2 }={ \omega  }^{ 3 }=1$

Hence equation whose roots are ${ \alpha  }^{ 19 },{ \beta  }^{ 7 }$ is

${ x }^{ 2 }-\left( { \alpha  }^{ 19 }+{ \beta  }^{ 7 } \right) x+{ \alpha  }^{ 19 }{ \beta  }^{ 7 }=0$

$\Rightarrow { x }^{ 2 }+x+1$

Which of the following quadratic equation has the sum of their roots $4$ and the sum of the cubes of their roots as $28$? 

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

  1. $x^2 - 4x + 3 = 0$

  2. $x^2 - 4x - 5 = 0$

  3. $x^2 - 3x + 4 = 0$

  4. $x^2 + 4x + 3 = 0$

Show answer
Correct Option: A
Explanation:

Let $\alpha$ and $\beta$ be the roots of the equation.
Hence
$\alpha+\beta=4$
and $\alpha^{3}+\beta^{3}=28$
Now $\alpha^{3}+\beta^{3}$ can be written as

$=(\alpha+\beta)^{3}-3\alpha\beta(\alpha+\beta)$
Hence
$28=64-12\alpha\beta$
$12\alpha\beta=36$
$\alpha\beta=3$
Therefore,
$x^{2}-(\alpha+\beta)x+\alpha\beta=0$
$x^{2}-(4x)+3=0$
Hence, option $A$ is correct.

If $\alpha$ and $\beta$ are the roots of the equation $4x^{2}\, -\, 5x\, +\, 2\, =\, 0$, find the equation whose roots are
$\displaystyle \frac{\alpha^{2}}{\beta}$ and $\displaystyle \frac{\beta^{2}}{\alpha}.$

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

  1. $2x^{2}\, -\, 5x\, +\, 16\, =\, 0$

  2. $32x^{2}\, +\, 5x\, +\, 16\, =\, 0$

  3. $2x^{2}\, +\, 5x\, +\, 16\, =\, 0$

  4. $32x^{2}\, -\, 5x\, +\, 16\, =\, 0$

Show answer
Correct Option: D
Explanation:

The given equation is: $4x^2 - 5x + 2 = 0 $


Sum of the roots = $\dfrac{5}{4}$

Product of the roots  = $\dfrac{2}{4} = \dfrac{1}{2}$

If the roots are $\dfrac{\alpha^2}{\beta}, \dfrac{\beta^2}{\alpha}$

Sum of roots = $\dfrac{\alpha^2}{\beta} + \dfrac{\beta^2}{\alpha}$ 

= $\dfrac {\alpha^3 + \beta^3}{\alpha\beta}$

= $\dfrac{{(\alpha + \beta)}^3 - 3\alpha\beta(\alpha+\beta)}{\alpha\beta}$

= $\dfrac{\left (\dfrac{5}{4}\right )^3 - 3 \left (\dfrac{5}{4}\right )\left (\dfrac{1}{2} \right )}{\dfrac{1}{2}}$

= $\dfrac{125 - 120}{32}$

= $\dfrac{5}{32}$

Product of roots = $ \left (\dfrac{\alpha^2}{\beta} \right ) \left ( \dfrac{\beta^2}{\alpha} \right )$

= $\alpha\beta$

= $\dfrac{1}{2}$

Hence.the equation in the standard form, $x^2 - Sx + P  = 0$ can be written as:

=$x^2 - \dfrac{5}{32}x + \dfrac{1}{2} = 0$

= $32x^2 - 5x + 16 = 0$

If $\alpha$ and $\beta$ are the roots of the equation $4x^{2}\, -\, 5x\, +\, 2\, =\, 0$, find the equation whose roots are
$\alpha\, +\, 3\beta$ and $3\alpha\, +\, \beta$.

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

  1. $16x^{2}\, +\, 80x\, +\, 107\, =\, 0$

  2. $16x^{2}\, -\, 80x\, +\, 107\, =\, 0$

  3. $16x^{2}\, -\, 80x\, -\, 107\, =\, 0$

  4. $16x^{2}\, +\, 80x\, -\, 107\, =\, 0$

Show answer
Correct Option: B
Explanation:

$4x^2 - 5x + 2 = 0 $


If $\alpha$ and $\beta$ are the roots of this equation, 

then , sum of roots: $\alpha + \beta$ = $\displaystyle \frac{5}{4}$

Product of roots: $\alpha. \beta = \displaystyle \frac{2}{4}$

The equation which has roots as : $\alpha + 3\beta$ and $\beta + 3\alpha$

Sum of roots: $4\alpha + 4\beta$ = $4 \left (\dfrac{5}{4} \right ) = 5$

Product of roots: $(\alpha + 3\beta)(3\alpha + \beta) $

$= 3(\alpha^2 + \beta^2) + 10\alpha\beta$

$= 3(\alpha + \beta)^2 - 6\alpha\beta + 10\alpha\beta$

$= 3 \left (\dfrac{5}{4} \right )^2 + 4\frac{2}{4}$

$= \dfrac{107}{16}$

Thus new equation is :$x^2 -Sx + P = 0$


$\therefore x^2 - 5x + \dfrac{107}{16} = 0$

$\therefore 16x^2 - 80x + 107 = 0$

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.

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

  1. Yes

  2. No

  3. Ambiguous

  4. Data insufficient

Show answer
Correct Option: A
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 $
$b^3 + a^2c + ac^2 = 3abc$

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

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

  1. $a = 2, b = 3$

  2. $a = 2, b = -3$

  3. $a = 5, b = -3$

  4. $a = 5, b = 3$

Show answer
Correct Option: C
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 $5{ x }^{ 2 }-3x+2=0$.

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

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

  1. ${x}^{2}+3{x}-15=0$

  2. ${x}^{2}+3{x}+73=0$

  3. ${x}^{2}+19{x}+73=0$

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

Show answer
Correct Option: C
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 .$

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 $

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

  1. $\displaystyle x^{2}=1$

  2. $\displaystyle x^{2}=4$

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

  4. None of these

Show answer
Correct Option: C
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 $

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

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

  1. $-4$

  2. $\dfrac{23}{6}$

  3. $13$

  4. $-8$

Show answer
Correct Option: A
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 $(a+b+g) =\displaystyle \frac{-b}{a}=\frac{-4}{1}=-4$