Tag: sum and product of the roots of a polynomial equation

Questions Related to sum and product of the roots of a polynomial equation

One root is three times the other, find the condition for a general quadratic equation

maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

  1. $\displaystyle 3b^{2}= 16ac$

  2. $\displaystyle 3b^{2}= ac$

  3. $\displaystyle b^{2}= 16ac$

  4. $\displaystyle 9b^{2}= 16ac$

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Correct Option: A
Explanation:

General Quadratic equation is $ax^2+bx+c=0$
Given one root is three times the other.
i.e $\alpha,3\alpha$ are the roots.
Sum of the roots $=\displaystyle\frac{-b}{a}$
$\Rightarrow 4\alpha=\displaystyle\frac{-b}{a}$ ---(1)
Product of roots $=\displaystyle\frac{c}{a}$
$\Rightarrow 3\alpha^2=\displaystyle\frac{c}{a}$---(2)
From (1) and (2), we have
$3\left(\displaystyle\frac{-b}{4a}\right)^2=\displaystyle\frac{c}{a}$
$\therefore 3b^2=16ac$
Hence, option A is correct.

Roots of the equation $\displaystyle (x+1)(x+2)(x+2)(x+3)(x+6)=15x^{2}$ are

maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

  1. all real & rational

  2. all non real

  3. two rational and two imaginary

  4. two imaginary and two irrational

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Correct Option: D

If one root of $x^{3}+ax^{2}+bx+c=0$ is the sum of the other two roots, then

maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

  1. $a^{3}=4(ab-c)$

  2. $a^{3}=4(ab-2c)$

  3. $a^{3}=ab-c$

  4. $a^{3}=ab-2c$

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Correct Option: B
Explanation:

Let the roots be $\alpha,\beta,\gamma$
Then
$\alpha=\beta+\gamma$.
Hence
$\alpha+\beta+\gamma=-a$
$2(\beta+\gamma)=-a$
$\beta+\gamma=\alpha=\dfrac{-a}{2}$ ...(i)
$\alpha.\beta+\beta.\gamma+\gamma.\alpha=b$
$\alpha(\beta+\gamma)+\beta.\gamma=b$
$\alpha^{2}+\beta.\gamma=b$
Or 
$\dfrac{a^{2}}{4}+\beta.\gamma=b$
$a^{2}+4\beta.\gamma=4b$ ...(ii)
And 
$\alpha.\beta.\gamma=-c$
Or 
$\dfrac{-a}{2}.\beta.\gamma=-c$
Or 
$\beta.\gamma=\dfrac{2c}{a}$.
Then
$a^{2}+4\beta.\gamma=4b$
$a^{2}+4\dfrac{2c}{a}=4b$
$a^{3}+8c=4ab$
Or 
$a^{3}=4(ab-2c)$.

If the sum of two roots of the equation $x^{3}-3x^{2}+kx+48=0$ is zero, then $k=$

maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

  1. $16$

  2. $-16$

  3. $24$

  4. $-24$

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Correct Option: B
Explanation:

Sum of two roots is zero. Using this we get the other root as $3$
Substitute $x=3$ and equating to zero we get, $k=-16$

One root of $x^{3}+x^{2}-2x-1=0$ lies between 

maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

  1. $-1$ and $0$

  2. $-2$ and $-1$

  3. $-3$ and $-2$

  4. $-4$ and $-3$

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Correct Option: A,B
Explanation:

$x^{ 3 }+x^{ 2 }-2x-1=0$
$f\left( 0 \right) =-1\ f\left( -1 \right) =1\ f\left( -2 \right) =-1\ f\left( -3 \right) =-13\ f\left( -4 \right) =-71$
As $f\left( -1 \right) >0$ and $f\left( -2 \right) <0$
Therefore one roots lies between $-2$ and $-1$

If two roots $\alpha,\beta$ of the equation $x^{4}-5x^{3}+11x^{2}-13x+6=0$ are connected by the relation $2\alpha+3\beta=7$, then the roots of the equation are

maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

  1. $-1,3,1\pm i\sqrt{2}$

  2. $-1,3,1\pm i\sqrt{3}$

  3. $2, 1,1\pm i\sqrt{2}$

  4. $2, 1,1\pm i\sqrt{3}$

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Correct Option: C
Explanation:

Let $\alpha ,\beta ,\gamma ,\delta $ are roots of $x^{ 4 }-5x^{ 3 }+11x^{ 2 }-13x+6=0$
${ s } _{ 1 }=\alpha +\beta +\gamma +\delta =5\ { s } _{ 4 }=\alpha \beta \gamma \delta =6$


For $\gamma ,\delta =1\pm i\sqrt { 2 } $ or $1\pm i\sqrt { 3 } \quad $
${ s } _{ 1 }\Rightarrow \alpha +\beta +2=5\Rightarrow \alpha +\beta =3$
Solving this with $2\alpha +3\beta =7$ we get
$\alpha =2$ and $\beta =1$

Now for $\gamma ,\delta =1\pm i\sqrt { 2 } $
$\alpha \beta \gamma \delta =2\left( 1+2 \right) =6$

And for $\gamma ,\delta =1\pm i\sqrt { 3 } $
$\alpha \beta \gamma \delta =2\left( 1+3 \right) =8$, not possible

Therefore, roots are $2, 1, 1\pm\sqrt{2}$
Hence, option 'C' is correct.

lf the difference of the squares of the roots of equation ${x}^{2} -6x+q=0$ is $24$, then the value of ${q}$ is:

maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

  1. $ -7$

  2. $8$

  3. $5$

  4. $4$

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Correct Option: C
Explanation:

Let $\alpha,\beta$ are roots of ${x}^{2}-6x+q=0,$ then

${ S } _{ 1 }=\alpha +\beta =6$
And ${ S } _{ 2 }=\alpha \beta =q$

Given ${ \alpha  }^{ 2 }-{ \beta  }^{ 2 }=24$
Now from ${ \left( \alpha -\beta  \right)  }^{ 2 }={ \left( \alpha +\beta  \right)  }^{ 2 }-4\alpha \beta $
$\Rightarrow { \left( \alpha -\beta  \right)  }^{ 2 }=36-4q\Rightarrow \left( \alpha -\beta  \right) =\sqrt { 36-4q } $

As ${ \alpha  }^{ 2 }-{ \beta  }^{ 2 }=24\Rightarrow \left( \alpha -\beta  \right) \left( \alpha +\beta  \right) =24$
$\Rightarrow \sqrt { 36-4q } \left( 6 \right) =24\Rightarrow \sqrt { 36-4q } =4$
$\Rightarrow 36-4q=16\Rightarrow 4q=20\Rightarrow q=5$

If the equation $\mathrm{a} _{\mathrm{n}}\mathrm{x}^{\mathrm{n}}+\mathrm{a} _{\mathrm{n}-1}\mathrm{x}^{\mathrm{n}-1}+\ldots\ldots+\mathrm{a} _{1}\mathrm{x}=0,\ \mathrm{a} _{1}\neq 0,\ \mathrm{n}\geq 2$, has a positive root $\mathrm{x}=\alpha$, then the equation $\mathrm{n}\mathrm{a} _{\mathrm{n}}\mathrm{x}^{\mathrm{n}-1}+(\mathrm{n}-1)\mathrm{a} _{\mathrm{n}-1}\mathrm{x}^{\mathrm{n}-2}+\ldots..+\mathrm{a} _{1}=0$ has a positive root, which is 

maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

  1. greater than $\alpha$

  2. smaller than $\alpha$

  3. greater than or equal to $\alpha$

  4. equal to $\alpha$

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Correct Option: B
Explanation:

=$ \because { a } _{ n }{ x }^{ 2\  }+{ a } _{ n }+{ x }^{ n-1 }+............+{ a } _{ 1 }x=\quad 0\quad \quad \quad { a } _{ 1 }\neq 0\quad n\ge 2 $

= has the root $x=\infty$ 
= ${ f }^{ 1 }(x)=\quad x{ a } _{ n }{ x }^{ n-1 }+\quad (x-1)\quad { a } _{ n-1 }{ x }^{ n-2 }+.......{ a } _{ n }$
= $\because f(x)=0$
Let us take an example to see 
Let a quadratic equation ${ x }^{ 2 }+2x-3=0$
${ x }^{ 2 }+3x-x-3=0$
$x(x+3)-1(x+3)=0 ........(i)$
$x=1\quad x=-3$
Now ${ f }^{ 1 }(x)=\quad 2x+1$
${ f }^{ 1 }(x)=\quad 0\quad =>\quad x=\quad -\cfrac { 1 }{ 2 } ..........(ii) $
From (i) and (ii) we can see that
The root of ${ f }^{ 1 }(x)$ is always less than the root of $f(x)$
Hence we can conclude
for $n{ a } _{ n }{ x }^{ n-1 }+(n-1){ a } _{ n-1 }{ x }^{ n-2 }+......{ a } _{ 1 }$
has roots always less than $\alpha $ for the value of $\alpha$.


If the sum of two roots of $x^{3}+ax+b=0$ is zero, then the value of $b$, is:

maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

  1. $a$

  2. $1$

  3. $-1$

  4. $0$

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Correct Option: D
Explanation:

$let\quad \alpha ,-\alpha ,\beta \quad be\quad the\quad roots\ Given\quad sum\quad of\quad the\quad roots\quad is\quad zero\ \alpha -\alpha +\beta =0\ \beta =0\ Therefore\quad product\quad of\quad the\quad roots\quad is\quad zero,\quad i.e.,\quad b=0$

lf one root of $\mathrm{x}^{2}-\mathrm{x}-\mathrm{k}=0(\mathrm{k}>0)$ is the square of the other root, then $\mathrm{k}=$ 

maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

  1. $ 2\pm\sqrt{5}$

  2. $ 2+\sqrt{5}$

  3. $ 2-\sqrt{5}$

  4. $1$

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Correct Option: B
Explanation:

$ Let\quad \alpha \quad and\quad \alpha ^{ 2 }\quad be\quad the\quad roots\quad as\quad per\quad the\quad given\quad condition.\ \therefore \quad \alpha +\alpha ^{ 2 }=1\quad and\quad { \alpha  }^{ 3 }=-k\ Now\quad (\alpha +\alpha ^{ 2 })^{ 3 }=1\ \Rightarrow { \alpha  }^{ 3 }+({ \alpha  }^{ 2 })^{ 3 }+3{ \alpha  }^{ 3 }(\alpha +\alpha ^{ 2 })=1\ Replacing\quad { \alpha  }^{ 3 }\quad by\quad -k\quad we\quad get\quad \ -k+k^{ 2 }-3k-1=0\ \Rightarrow { k }^{ 2 }-4k-1=0\ \Rightarrow k=\frac { 4+\sqrt { 20 }  }{ 2 } =2+\sqrt { 5 } \ or\quad k=\frac { 4-\sqrt { 20 }  }{ 2 } =2-\sqrt { 5 } <0\ But\quad k>0\quad therefore\quad we\quad reject\quad this\quad value.\ \therefore \quad k=2+\sqrt { 5 } \ Answer-\quad Option\quad B. $