Tag: descartes rule

Questions Related to descartes rule

Suppose $f(x) =3x^3-13x^2+14x-2$, it is assumed that $f(x)=0$ will have 3 root say $\alpha, \beta$ and $\gamma$, where $\alpha < \beta < \gamma$

$[\alpha], [\beta], [\gamma]$ (where, [-] denotes the greatest function) will be in

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

  1. AP

  2. GP

  3. HP

  4. None of these

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Correct Option: A
Explanation:

$f\left( x \right) =3{ x }^{ 3}-13{ x }^{ 2 }+14x-2$


$f\left( 0 \right) =-2=-ve$

$f\left( 1 \right) =3-13+14-2=2=+ve$

$f\left( 2 \right) =24-52+28-2=-2=-ve$

$f\left( 3 \right) =81-117+42-2=4=+ve$

$\therefore $ One root lies between 0 & 1

One root lies between 1 & 2

One root lies between 2 & 3

$\therefore \alpha \in \left( 0,1 \right) \Rightarrow \left[ \alpha  \right] =0$

$\beta \in  \left( 1,2 \right) \Rightarrow \left[ \beta  \right] =1$

$\gamma \in \left( 2,3 \right) \Rightarrow \left[ \gamma  \right] =2$

$\left[ \alpha  \right] ,\left[ \beta  \right] ,\left[ \gamma  \right] $ are in AP

lf one root of the equation $ax^{2}+bx+c=0$ is the square of the other, then

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

  1. $b^{2}+ac^{2}+a^{2}c=3abc$

  2. $b^{3}+ac^{2}+a^{2}c=3abc$

  3. $b^{2}+ac^{2}+a^{2}c+3abc=0$

  4. $b^{3}+ac^{2}+a^{2}c+3abc=0$

Show answer
Correct Option: B
Explanation:

Given equation $a{ x }^{ 2 }+bx+c$
Given that, one root of the equation is square of another.
So, lets assume $\alpha$ , ${ \alpha  }^{ 2 }$ are roots of the given equation
We know that,
Sum of roots $=$ $\alpha +{ \alpha  }^{ 2 }=\dfrac { -b }{ a }$ 
Product of roots $=$ $ \alpha \times { \alpha  }^{ 2 }=\dfrac { c }{ a }$
$\alpha (1+\alpha )=\dfrac { -b }{ a } \longrightarrow 1  $
${ \alpha  }^{ 3 }=\dfrac { c }{ a } \longrightarrow 2 $
Cubing equation (1) on both sides and substitute the value from equation (2).
${ \alpha  }^{ 3 }{ (1+\alpha ) }^{ 3 }=\dfrac { -{ b }^{ 3 } }{ { a }^{ 3 } } \ { \alpha  }^{ 3 }({ \alpha  }^{ 3 }+1+3{ \alpha  }(1+\alpha ))=\dfrac { -{ b }^{ 3 } }{ { a }^{ 3 } } \ \dfrac { c }{ a } \left (\dfrac { c }{ a } +1+3\left (\dfrac { -b }{ a } \right)\right)=\dfrac { -{ b }^{ 3 } }{ { a }^{ 3 } } \ \dfrac { ({ c }^{ 2 }+ac-3bc) }{ { a }^{ 2 } } =\dfrac { -{ b }^{ 3 } }{ { a }^{ 3 } } \ a({ c }^{ 2 }+ac-3bc)=-{ b }^{ 3 }\ { b }^{ 3 }+a{ c }^{ 2 }+{ a }^{ 2 }c=3abc $