Tag: theory of equations

Questions Related to theory of equations

Rohan and Sohan were attempting to solve the quadratic equation  $\displaystyle x^{2}-ax+b=0$. Rohan copied the coefficient of x wrongly and obtained the roots as 4 and 12 . Sohan copied the constant term wrongly and obtained the roots as -19 and 3. Find the correct roots

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

  1. -8, -10

  2. -8, -6

  3. -4, -12

  4. 4, 12

Show answer
Correct Option: C
Explanation:

With the roots $ 4, 12 $, the equation was $ (x-4))(x-12) ={x}^{2} -4x  -12x + 48 = {x}^{2} -16x + 48 $

As Rohan made a mistake in noting the coffecient of $ x $ , in the original equation, coefficient of $ {x}^{2} = 1 $ and constant $ = 48 $

Now, with the roots $ -19, 3 $, the equation was $ (x-(-19))(x-3) = (x+19)(x-3) = {x}^{2} + 19x -3x -57 = {x}^{2} + 16x -57 $

As Sohan made a mistake in noting just the constant term in the original  equation, coefficient of $ {x}^{2} = 1 $ and of $ x = 16 $

So, we get the original equation as $ {x}^{2} + 16x + 48 = 0 $
Solving it, we get $ {x}^{2} + 4x +12x + 48 = 0 $
$ => x(x+4) + 12(x+4) = 0 $
$ => (x+4)(x+12) = 0 $
$ => x = -4, -12 $

If the equation formed by decreasing each root of $ax^{2}+bx+c=0$ by $1$ is $2x^{2}+8x+2=0$, then

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

  1. $\mathrm{a}=-\mathrm{b}$

  2. $\mathrm{b}=-\mathrm{c}$

  3. $\mathrm{c}=-\mathrm{a}$

  4. $\mathrm{b}=\mathrm{a}+\mathrm{c}$

Show answer
Correct Option: B
Explanation:

Since the equation $2{ x }^{ 2 }+8x+2=0$ has roots which are 1 less than those of the equation $a{ x }^{ 2 }+bx+c=0$ then if we replace $ x $ by $ x+1 $ in latter we'll get the former.
$\Rightarrow a(x+1)^{ 2 }+b(x+1)+c=0$
$\Rightarrow  a{ x }^{ 2 }+(2a+b)x+a+b+c=0$
comparing this equation with that of $2{ x }^{ 2 }+8x+2=0$
we get option (b) as the correct answer

If $\displaystyle \alpha ,\beta $ are the roots of $\displaystyle x^{2}+x+1=0 $ and $\displaystyle \gamma ,\delta  $ are the roots of $\displaystyle x^{2}+3x+1=0 $ then $\displaystyle \left ( \alpha -\gamma  \right )\left ( \beta +\delta  \right )\left ( \alpha +\delta  \right )\left ( \beta -\gamma  \right )$ = 

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

  1. 2

  2. 4

  3. 6

  4. 8

Show answer
Correct Option: D
Explanation:
${ x }^{ 2 }+x+1=0$
If $\alpha ,\beta $ are the roots, then $\alpha +\beta =-1\quad ,\quad \alpha \beta =1$
Solving the above equation, we get the $\alpha =\dfrac { -1+i\sqrt { 3 }  }{ 2 } ,\quad \beta =\dfrac { -1-i\sqrt { 3 }  }{ 2 } \\ $

Similarly, ${ x }^{ 2 }+3x+1=0$
If $\gamma ,\delta  $ are the roots, then $\gamma +\delta =-3\quad ,\quad \gamma \delta =1$
Solving the above equation, we get the $\gamma =\dfrac { -3+\sqrt { 5 }  }{ 2 } ,\quad \delta =\dfrac { -3-\sqrt { 5 }  }{ 2 } $

$(\alpha -\gamma )(\beta -\gamma )(\alpha +\delta )(\beta +\delta )$ is..
$(\alpha -\gamma )(\beta -\gamma )$
$=\alpha \beta -\gamma (\alpha +\beta )+{ \gamma  }^{ 2 }\\ =1-(-1)(\dfrac { -3+\sqrt { 5 }  }{ 2 } )+(\dfrac { 14-6\sqrt { 5 }  }{ 4 } )\\ =1+(\dfrac { -3+\sqrt { 5 }  }{ 2 } )+(\dfrac { 14-6\sqrt { 5 }  }{ 4 } )\\ =\dfrac { 12-4\sqrt { 5 }  }{ 4 } \\ =3-\sqrt { 5 } $

$(\alpha +\delta )(\beta +\delta )\\ =\alpha \beta +\delta (\alpha +\beta )+\delta ^{ 2 }\\ =1+(-1)(\dfrac { -3-\sqrt { 5 }  }{ 2 } )+(\dfrac { 14+6\sqrt { 5 }  }{ 4 } )\\ =1+(\dfrac { 3+\sqrt { 5 }  }{ 2 } )+(\dfrac { 14+6\sqrt { 5 }  }{ 4 } )\\ =\dfrac { 24+8\sqrt { 5 }  }{ 4 } \\ =6+2\sqrt { 5 } =\quad 2(3+\sqrt { 5 } )$

Multiplying the above two results, we get
$2(3+\sqrt { 5 } )(3-\sqrt { 5 } )\\ =\quad 2(9\quad -\quad 5)\quad \\ =8$


Umesh and Varun are solving an equation of the form $\displaystyle x^{2}+bx+c=0$. In doing so Umesh commits a mistake in noting down the constant term and finds the roots as $-3$ and $-12$. And Varun commits a mistake in noting down the coefficient of $x$ and find the roots as $-27$ and $-2$. If so find the original equation

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

  1. $\displaystyle x^{2}-15x+36=0$

  2. $\displaystyle x^{2}+15x+36=0$

  3. $\displaystyle x^{2}-15x+54=0$

  4. $\displaystyle x^{2}+15x+54=0$

Show answer
Correct Option: D
Explanation:

With the roots $ -3, -12 $, the equation was $ (x-(-3))(x-(-12) = (x+3)(x+6) = {x}^{2} + 3x + 12x + 36 = {x}^{2} + 15x + 36 $

As Umesh made a mistake in noting just the constant term, in the original equation, coefficient of $ {x}^{2} = 1 $ and of $ x = 15 $

Now, with the roots $ -27, -2 $, the equation was $ (x-(-27))(x-(-2)) = (x+27)(x+2) = {x}^{2} + 27x + 2x + 54 = {x}^{2} + 29x + 54 $

As Varun made a mistake in noting the coffecient of $ x $ in the original  equation, coefficient of $ {x}^{2} = 1 $ and constant $ = 54 $

So, we get the original equation as $ {x}^{2} + 15x + 54 = 0 $