Tag: properties of roots of a quadratic equations

Questions Related to properties of roots of a quadratic equations

If the difference of the roots of the quadratic equation is 3 and difference between their cubes is 189, then the quadratic equation is x2±9x+18=0x2±9x+18=0
State true or false.

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

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

Let the roots of the equation be a and b
then, $a^3 - b^3 = 189$
and $a - b = 3$
cubing both sides:
$(a-b)^3 = 27 $
$a^3 - b^3 - 3ab (a-b) = 27$
$189 -3ab(3) = 27 $
$162 = 9 ab$
$ab = 18$
Similarly, $(a+b)^2 = (a -b)^2  + 4ab$
$(a+b)^2 = 3^2 + 4(18)$
$(a+b)^2 = 9 + 72 $
$a +b = \pm 9$
The general form of equation is $x^2 -Sx + P = 0 $, hence the equation will be
$x^2 \pm 9x + 18 = 0$

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

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

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

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

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

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

Show answer
Correct Option: A,D
Explanation:

If $\alpha, \beta$ are the roots of $x^2-2x+3=0$
then $\displaystyle \alpha ^{2}-2\alpha +3= 0$ ...(1)
and $\displaystyle \beta^2-2\beta+3=0$  ....(2)
$\displaystyle \therefore \alpha ^{2}= 2\alpha -3, \alpha ^{3}= 2\alpha ^{2}-3\alpha $
$\displaystyle \therefore P= \left ( 2\alpha ^{2}-3\alpha  \right )-3\alpha ^{2}+5\alpha -2$
$\displaystyle = -\alpha ^{2}+2\alpha -2= 3-2= 1,$ by (1)
Similarly $\displaystyle Q= 2 \therefore S= 3, P= 2$
Hence reqd. eq. is $\displaystyle x^{2}-3x+2= 0.$ or $-x^2+3x-2=0$

If the difference of the roots of a quadratic equation is 4 and the difference of their cubes is 208, then the quadratic equation is $x^{2}\, \pm\, 8x\, +\, 12\, =\, 0$
State true or false.

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

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

Let the roots of the equation be a and b
then, $a^3 - b^3 = 208$
and $a - b = 4$
cubing both sides:
$(a-b)^3 = 64 $
$a^3 - b^3 - 3ab (a-b) = 64$
$208 -3ab(4) = 64 $
$144 = 12 ab$
$ab = 12$
Similarly, $(a+b)^2 = (a -b)^2  + 4ab$
$(a+b)^2 = 4^2 + 4(12)$
$(a+b)^2 = 16 + 48 $
$a +b = \pm 8$
The general form of equation is $x^2 -Sx + P = 0 $, hence the equation will be
$x^2 \pm 8x + 12 = 0$

If $\alpha$ and $\beta$ are the roots of the equation $4x^{2}\, -\, 5x\, +\, 2\, =\, 0$, find the equation whose roots are
$\alpha\, +\, \displaystyle \frac{1}{\alpha}$ and $\beta\, +\, \displaystyle \frac{1}{\beta}$.

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

  1. $8x^{2}\, +\, 30x\, +\, 29\, =\, 0$

  2. $x^{2}\, -\, 30x\, +\, 29\, =\, 0$

  3. $8x^{2}\, -\, 30x\, +\, 29\, =\, 0$

  4. $x^{2}\, +\, 30x\, +\, 29\, =\, 0$

Show answer
Correct Option: C
Explanation:

The 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 $\alpha + \dfrac{1}{\alpha}, \beta + \dfrac{1}{\beta}$

Sum of roots = $\alpha + \dfrac{1}{\alpha} + \beta + \dfrac{1}{\beta}$
 
= $\alpha + \beta + \dfrac{\alpha+ \beta}{\alpha\beta}$

= $\dfrac{5}{4} + \dfrac{\dfrac{5}{4}}{\dfrac{1}{2}}$

= $\dfrac{5}{4} + \dfrac{5}{2}$

$\dfrac{15}{4}$

Product of roots = $(\alpha + \dfrac{1}{\alpha})( \beta + \dfrac{1}{\beta})$

= $\alpha\beta + \dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} + \dfrac{1}{\alpha\beta}$

= $\dfrac{1}{2} + \dfrac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta} + 2$

= $\dfrac{1}{2} + \dfrac{\dfrac{25}{16} - 1}{\dfrac{1}{2}} + 2$

= $\dfrac{1}{2} + \dfrac{9}{8} + 2$

= $\dfrac{4 + 9 + 16}{8}$ 

= $\dfrac{29}{8}$

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

=$x^2 - \dfrac{15}{4}x + \dfrac{29}{8} = 0$

= $8x^2 - 30x + 29 = 0$

Hence option $'C'$ is the answer.

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