Tag: descartes rule

Questions Related to descartes rule

If the sum of two roots of the equation $\displaystyle x^{3}+ax^{2}+bx+c= 0 $ is zero, then value of $ab$ equals

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

  1. $c$

  2. $2c$

  3. $-2c$

  4. $-c$

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Correct Option: A
Explanation:

Given equation is $x^{3}+ax^{2}+bx+c= 0$

Let the roots be $\alpha, -\alpha, \beta$

Then $\alpha-\alpha+\beta=-a$

$\Rightarrow \beta=-a$       ....(1)

Also, $-{\alpha}^{2}+{\alpha}\beta-\alpha\beta=b$

$\Rightarrow -{\alpha}^{2}=b$       .....(2)

Also, $-{\alpha}^{2} \beta=-c$  ....(by (1)and (2))

$\Rightarrow -ab=-c$

$\displaystyle \Rightarrow ab= c $ 

If $\displaystyle x^{3}-mx^{2}-3x+2=0$ has two roots equal in magnitude but opposite in sign, then $m$ is:

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

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

  2. $\displaystyle \frac{2}{3}$

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

  4. none of these

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Correct Option: B
Explanation:

Let $\alpha ,-\alpha ,\beta $ be the roots of $x^{ 3 }-mx^{ 2 }-3x+2=0$
Then
${ s } _{ 1 }=\alpha -\alpha +\beta =m\ \Rightarrow \beta =m$
Substituting $x=m$ in equation, we get
$m^{ 3 }-m.m^{ 2 }-3.m+2=0\ \Rightarrow m=\cfrac { 2 }{ 3 } $
Hence, option 'B' is correct.

The equation $\displaystyle x^{4} - x^{3} + 1 = 0$, has

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

  1. all imaginary roots

  2. all four real roots

  3. two real and two imaginary roots

  4. none of these

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Correct Option: C
Explanation:

Let $f(x)=x^4-x^3+1$

our first case is the positive-root case: 
In $f(x),$ there are two sign changes in the positive-root case. 
This number "two" is the maximum possible number of positive zeroes (that is, all the positive x-intercepts) for the given polynomial.
I've finished the positive-root case, so now I look at $f(-x)$. That is, having changed the sign on $x$, I'm now doing the negative-root case:
$f(-x)=x^4+x^3+1$
There is zero sign change in this negative-root case, so there is no negative root.
Therefore, there are max two positive roots and no negative roots, therefore remaining two roots are imaginary.
Hence, option C is correct. 

The equation $x-\dfrac{2}{x-1}=1-\dfrac{2}{x-1}$ has

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

  1. no root

  2. one root

  3. two equal roots

  4. infinitely many roots

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Correct Option: A
Explanation:

For equation $x-\dfrac { 2 }{ x-1 } =1-\dfrac { 2 }{ x-1 } $, the term $'x-1'$ is in the denominator. Hence the solution isn't defined. For $x=1$ $\Rightarrow $ $x\neq 1$

We have our equation as $x-\dfrac { 2 }{ x-1 } =1-\dfrac { 2 }{ x-1 } $
cancelling the common term on both sides,we get $x=1$. 
But for well defined solution $x\neq 1$. Hence,this equation has no solution.

The equation $\displaystyle x - \frac{5}{x - 2} = 2 - \frac{5}{x - 2}$ has

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

  1. No real roots

  2. Only one real root

  3. Two real roots

  4. Infinitely many roots

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Correct Option: A
Explanation:

$x-\dfrac { 5 }{ x-2 } =2-\dfrac { 5 }{ x-2 } $       ...(1)
Equation (1) is valid when $x\neq 2$
Rewriting eq. (1), we get $x=2$
But $x\neq 2$
Therefore, number of roots satisfying eq. (1) are zero.

Number of real roots of equation $\displaystyle 2x^{99}+3x^{98}+2x^{97}+3x^{96}+........+2x+3=0$ are

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

  1. $99$

  2. $49$

  3. $1$

  4. $3$

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Correct Option: C
Explanation:

 $2x^{99}+3x^{98}+2x^{97}+3x^{96}+........+2x+3=0$ .... $(i)$

Taking $2x+3$ common, we get
$(2x+3)(x^{98}+x^{97}+...1)=0$
Since $x^{98}+x^{97}+...1$ can not be equal to zero
Therefore only $2x+3 =0$ or $x=-\dfrac{3}{2}$ is real root
Hence, number of real roots of equation $(i)$ is $1$.

The number of rational roots of $\displaystyle x^{10}-x^{9}-2=0$:

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

  1. $3$

  2. $2$

  3. $1$

  4. $0$

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Correct Option: C
Explanation:

$x^{10}-x^9-2=x^{10}-2x^9+1x^9-2=(x^9+1)(x-2)$

$x=2,x^9=-1\Rightarrow x=\sqrt[9]{-1}$
Hence it has only one rational root.

If the equation $\displaystyle 5x^{5}-25x^{4}+ax^{3}+bx^{2}+cx-5=0$ has five positive roots, then the value of $2a + 3b + 2c$ is 
maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

  1. 60

  2. 300

  3. 0

  4. cannot be determine

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Correct Option: C

Find the number of rational roots of 
$\displaystyle P(x)=2x^{98}+3x^{97}+2x^{96}+.....+2x+3=0$

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

  1. $2$

  2. $3$

  3. $4$

  4. $5$

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Correct Option: A
Explanation:

$P(x)=2x^{98}+3x^{97}+2x^{96}+.....+2x+3=0$

Carrying out the $2x+3$ and $x+1$ as common term
$(2x+3)(x^{97}+x^{96}+....1)=(2x+3)(x+1)(x^{96}+x^{94}+....1)$  
Hecne two rational roots are $x=\dfrac{-3}{2},-1$

The condition for the equation $\displaystyle ax^{2}+bx+c= 0$ to have one root $n$ times the other, is:

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

  1. $\displaystyle na^{2}= bc\left ( n+1 \right )^{2}$

  2. $\displaystyle nb^{2}= ac\left ( n+1 \right )^{2}$

  3. $\displaystyle nb^{2}= ac\left ( n-1 \right )^{2}$

  4. None of these

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Correct Option: B
Explanation:

Given that the roots of the equation $ ax^2+bx+c=0 $ be such that one root is $n$ times the other. 
Let one root be $\alpha$, then the other root will be $n\alpha$ by given condition.
Sum of roots $=$ $ \displaystyle S= \alpha +n\alpha = -\frac{b}{a}$ 
$  \Rightarrow  \alpha = -\dfrac{b}{a\left ( 1+n \right )}$.....(1)
Product of roots $=  n\alpha ^{2}= \dfrac{c}{a}$ 
$ \Rightarrow  \alpha ^{2}= \dfrac{c}{an}$ ....(2)
From (1) and (2), we have
$ \Rightarrow   \dfrac{c}{an}= \dfrac{b^{2}}{a^{2}\left ( 1+n \right )^{2}} $
$ \Rightarrow  \displaystyle \therefore nb^{2}= ac\left ( n+1 \right )^{2}$