Tag: vieta’s formula for quadratic equations

Questions Related to vieta’s formula for quadratic equations

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 $=\tan A\tan B$
$\angle C=90^o$
$\angle A+\angle B=90^o$
$\tan A \tan B=1$
$\tan B=1/\tan A$
$x^2-(\tan A+\tan B)x+1=0$
$x^2\dfrac {2}{\sin 2A}x+1=0$
$\Rightarrow \ x^2-\dfrac {c^2}{ab}x+1=0 \Rightarrow \ x^2 (ab)-c^2 (x)+1=0$
$\tan A+\tan B=\tan A+\dfrac {1}{\tan A}$
$=\dfrac {2\tan ^2 A+1}{2\tan A}=\dfrac {2}{\sin 2A}$
$\sin 2A=\dfrac {2\tan A}{1+\tan^2 A}$


If $\displaystyle \alpha $ are $\displaystyle \beta $ are the roots of $\displaystyle x^{2}+x+1=0$  then find the equation whose roots $\displaystyle \alpha ^{2}$ and $\displaystyle \beta ^{2}$

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

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

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

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

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

Show answer
Correct Option: A
Explanation:

For the given equation, sum of roots $ = \alpha + \beta  = -\dfrac {1}{1} = -1 $

Product of roots $ = \alpha \times \beta =  \dfrac {1}{1} =1 $

Now, $ {\alpha}^{2} + \beta ^{2} = (\alpha + \beta )^{2} - 2(\alpha \times \beta)= (-1)^{2} - 2(1) = 1-2 = -1 $

And $ {\alpha}^{2} \times \beta ^{2} = (\alpha \times \beta )^{2} = 1 $

Equation whose roots are $ {\alpha}^{2} $ and $ \beta ^{2} $ is $ x^{2} -(Sum \ of \ roots)x +  Product \ of \ roots  = 0 $
$ => x^{2} -({\alpha}^{2} + \beta ^{2})x +  {\alpha}^{2} \times \beta ^{2}  = 0 $
$ => x^{2} -(-1)x+ 1  = 0 $
$ => x^{2} +x+ 1  = 0 $

Two students Ragini and Gourav were asked to solve a quadratic equation $\displaystyle ax^{2}+bx+c=0,a\neq 0$ Ragini made some mistake in writing b and found the roots as 3 and $\displaystyle -\frac{1}{2}$ Gourav too made mistake in writing c and found the roots -1 and $\displaystyle -\frac{1}{4}$ The correct roots of the given equation should be

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

  1. $-2,$ $\displaystyle \frac{3}{4}$

  2. $3, -1$

  3. $\displaystyle -\frac{1}{2}$, -1

  4. $3,$ $\displaystyle -\frac{1}{4}$

Show answer
Correct Option: A
Explanation:

Given: Ragini found roots as $3, -\dfrac 12$ when copied the wrong coefficient of $x$ and Gourav found roots as $-1, -\dfrac 14$ when copied wrong constant term.

To find the correct roots of the given equation
Sol: Viete's formula for the roots $x _1$ and $x _2$ of equation $ax^2+bx+c=0$:  $x _1+x _2=−\dfrac ba$ and $ x _1\times x _2=\dfrac ca$
According to Ragini, she copied the constant term and the coefficient of $ x^2$ correctly. Hence $3\times -\dfrac 12=-\dfrac 32=\dfrac ca$
And according to Gourav, he copied coefficient of $x$ and $x^2$ correctly. Hence $-1+\left(-\dfrac 14\right)=-\dfrac 54=-\dfrac ba$
Hence the equation becomes,
$x^2+\dfrac 54x-\dfrac 32=0\implies 4x^2+5x-6=0$
$\implies 4x^2+8x-3x-6=0\\implies 4x(x+2)-3(x+2)=0\\implies (4x-3)(x+2)=0\\implies x _1=-2, x _2=\dfrac 34$
are the correct roots of the given equation.

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 $