Tag: theory of equations

Questions Related to theory of equations

If the roots of ${a _1}{x^2}\, + \,{b _1}x\, + \,{c _1}\, = \,0$ are ${\alpha _1},\,{\beta _1},\,$ and those of ${a _2}{x^2}\, + \,{b _2}x\, + {c _2}\, = \,0$ are ${\alpha _2}\,,{\beta _2}$ such that ${\alpha _1}\,{\alpha _2} = \,{\beta _1}\,{\beta _2}\, = \,1$, then

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

  1. $\dfrac{{{a _1}}}{{{a _2}}} = \,\dfrac{{{b _1}}}{{{b _2}}}\, = \,\dfrac{{{c _1}}}{{{c _2}}}$

  2. $\dfrac{{{a _1}}}{{{c _2}}} = \,\dfrac{{{b _1}}}{{{b _2}}}\, = \,\dfrac{{{c _1}}}{{{a _2}}}$

  3. ${a _1}\,{a _2}\, = \,{b _1}\,{b _2}\, = \,{c _1}\,{c _2}$

  4. None of these

Show answer
Correct Option: A
Explanation:

$a _1x^2+b _1x+c _1=0$

$\alpha _1+\beta _1=\cfrac{-b _1}{a _1}$
$\alpha _1\beta _1=\cfrac{c _1}{a _1}$

$a _2x^2+b _2x+c _2=0$
$\alpha _2+\beta _2=\cfrac{-b _2}{a _2}$
$\alpha _2\beta _2=\cfrac{c _2}{a _2}$

$\alpha _1\alpha _2=\beta _1\beta _2=1$
$\therefore \alpha _1\beta _1\alpha _2\beta _2=(\alpha _1\alpha _1)\cdot (\beta _1\beta _1)=1\cdot 1=1$
$\therefore \alpha _1\beta _1\alpha _2\beta _2=(\alpha _1\alpha _1)\cdot (\beta _1\beta _1)=\cfrac{c _1}{a _1}\cdot \cfrac{c _2}{a _2}=1$
$\implies \cfrac{c _1}{a _2}=\cfrac{a _1}{c _2}$
Now, $\alpha _1+\beta _1=\cfrac{-b _1}{a _1}$
$\implies \cfrac{1}{\alpha _2}+\cfrac{1}{\beta _2}=\cfrac{-b _1}{a _1}$
$\implies \cfrac{\alpha _2+\beta _2}{\alpha _2\beta _2}=\cfrac{-b _1}{a _1}$
$\implies \cfrac{-b _2/a _2}{c _2/a _2}=\cfrac{-b _1}{a _1}$
$\implies \cfrac{b _2}{c _2}=\cfrac{b _1}{a _1}$
$\implies \cfrac{a _1}{c _2}=\cfrac{b _1}{b _2}$
$\therefore \cfrac{a _1}{a _2}=\cfrac{b _1}{b _2}=\cfrac{c _1}{c _2}$

If $alpha, beta$ are roots of $Ax^2 + Bx + C = 0$ and $\alpha^2, \beta^2$ are roots of $x^2 + px + q = 0$, the $p$ is equal to

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

  1. $\dfrac{B^2 - 2AC}{A^2}$

  2. $\dfrac{2AC - B^2}{A^2}$

  3. $\dfrac{B^2 - 4AC}{A^2}$

  4. $\dfrac{4AC - B^2}{A^2}$

Show answer
Correct Option: B
Explanation:
$Ax^2+Bx+C=0$
$\alpha +\beta =\dfrac{-B}{A}$
$\alpha\beta =\dfrac{C}{A}$
Now it roots are $\alpha^2$ & $\beta^2$
Then equation will be
$(x-\alpha^2)(x-\beta^2)=0$
$x^2-x\beta^2-x\alpha^2+\alpha^2\beta^2=0$
$x^2-)(\alpha^2+\beta^2)x+\alpha^2\beta^2=0$
$x^2+[-(\alpha^2+\beta^2)]x+\alpha^2\beta^2=0$
$\alpha +\beta =\dfrac{-B}{-A}$
$\Rightarrow \alpha^2+\beta^2+2\alpha\beta =\dfrac{B^2}{A^2}$
$\alpha^2+\beta^2=\dfrac{B^2}{A^2}-\dfrac{2C}{A}$
$-(\alpha^2+\beta^2)=\dfrac{2C}{A}-\dfrac{B^2}{A^2}=\dfrac{2AC-B^2}{A^2}$
$p=\dfrac{2AC-B^2}{A^2}$.

If $\alpha+\beta$$=-2$ and ${\alpha}^{3}+{\beta}^{3}$$=-56$ then the quadratic equation whose roots are $\alpha,\beta$ is 

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

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

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

  3. ${x}^2+2x-12$$=0$

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

Show answer
Correct Option: D
Explanation:

$\Rightarrow$  $\alpha+\beta=-2$             ------ ( 1 )


$\Rightarrow$  $\alpha^3+\beta^3=-56$


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

$\Rightarrow$  $(\alpha+\beta)^3=\alpha^3+\beta^3+3\alpha\beta(\alpha+\beta)$

$\Rightarrow$  $(-2)^3=-56+3\alpha\beta(-2)$            [ Using ( 1 ) and ( 2 ) ]

$\Rightarrow$  $-8+56=-6\alpha\beta$

$\Rightarrow$  $48=-6\alpha\beta$

$\Rightarrow$  $\alpha\beta=-8$                      ----- ( 2 )

The required quadratic equation,

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

Using ( 1 ) and ( 3 ) we get,
$\Rightarrow$  $x^2+2x-8=0$

If $\alpha \neq \beta$ but $\alpha^2 = 5 \alpha -3$ and $\beta^2 = 5\beta -3$, then the equation whose roots are $\dfrac{\alpha}{\beta}$ and $\dfrac{\beta}{\alpha}$is

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

  1. $3x^2 - 25x+3=0$

  2. $x^2 +5x-3=0$

  3. $x^2 -5x+3=0$

  4. $3x^2 - 19x+3=0$

Show answer
Correct Option: D
Explanation:
${\alpha}^{2}=5\alpha-3$

${\alpha}^{2}-5\alpha+3=0$

$\alpha=\dfrac{5\pm\sqrt{{5}^{2}-4\times 1\times 3}}{2}$

$\alpha=\dfrac{5\pm\sqrt{25-12}}{2}$

$\alpha=\dfrac{5\pm\sqrt{13}}{2}$

${\beta}^{2}=5\beta-3$

${\beta}^{2}-5\beta+3=0$

$\beta=\dfrac{5\pm\sqrt{{5}^{2}-4\times 1\times 3}}{2}$

$\beta=\dfrac{5\pm\sqrt{25-12}}{2}$

$\beta=\dfrac{5\pm\sqrt{13}}{2}$

Given:$\alpha\neq\,\beta$

Let $\alpha=\dfrac{5+\sqrt{13}}{2}$ and $\beta=\dfrac{5-\sqrt{13}}{2}$

$\Rightarrow\,\dfrac{\alpha}{\beta}=\dfrac{\dfrac{5+\sqrt{13}}{2}}{\dfrac{5-\sqrt{13}}{2}}$

$=\dfrac{5+\sqrt{13}}{5-\sqrt{13}}$

$=\dfrac{5+\sqrt{13}}{5-\sqrt{13}}\times \dfrac{5+\sqrt{13}}{5+\sqrt{13}}$

$=\dfrac{25+13+10\sqrt{13}}{25-13}$

$=\dfrac{38+10\sqrt{13}}{12}$

$=\dfrac{19+5\sqrt{13}}{6}$

$\Rightarrow\,\dfrac{\beta}{\alpha}=\dfrac{\dfrac{5-\sqrt{13}}{2}}{\dfrac{5+\sqrt{13}}{2}}$

$=\dfrac{5-\sqrt{13}}{5+\sqrt{13}}$

$=\dfrac{5-\sqrt{13}}{5+\sqrt{13}}\times \dfrac{5-\sqrt{13}}{5-\sqrt{13}}$

$=\dfrac{25+13-10\sqrt{13}}{25-13}$

$=\dfrac{19-10\sqrt{13}}{12}$

$=\dfrac{19-5\sqrt{13}}{6}$

Sum of the zeroes$=\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}$

$=\dfrac{19+5\sqrt{13}}{6}+\dfrac{19-5\sqrt{13}}{6}$

$=\dfrac{19+5\sqrt{13}+19-5\sqrt{13}}{6}$

$=\dfrac{2\times 19}{6}=\dfrac{19}{3}$

Product of the zeroes$=\dfrac{\alpha}{\beta}\times\dfrac{\beta}{\alpha}$

$=\dfrac{19+5\sqrt{13}}{6}\times\dfrac{19-5\sqrt{13}}{6}$

$=\dfrac{361-25\times 13}{36}=\dfrac{361-325}{36}=\dfrac{36}{36}=1$

Now,we know the foumula for finding quadratic equations,
${x}^{2}-\left(sum\,of \,the \,zeroes\right)x+product\,of\,the \,zeroes=0$

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

Hence the equation is $3{x}^{2}-19x+3=0$

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

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

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