Tag: descartes rule

Questions Related to descartes rule

Given $P(x) = {x^4} + a{x^3} + b{x^2} + cx + d$ such that $x=0$ is the only real root of $P(x) = 0$. If $P(-1) < P(1) $,then in the interval $[-1,1]$

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

  1. $P(-1)$ is the minimum and $P(1)$ is the maximum of P

  2. $P(-1)$ is not the minimum but $P(1)$ is the maximum of P

  3. $P(-1)$ is the minimum and $P(1)$ is not the maximum of P

  4. neither $P(-1)$ is the minimum nor $P(1)$ is the maximum of P

Show answer
Correct Option: B
Explanation:
Given:$P\left(x\right)={x}^{4}+a{x}^{3}+b{x}^{2}+cx+d$

${P}^{\prime}\left(x\right)=4{x}^{3}+3a{x}^{2}+2bx+c$

Since, $x=0$ is a solution for ${P}^{\prime}\left(x\right)=0$

$\Rightarrow\,c=0$

So, $P\left(x\right)={x}^{4}+a{x}^{3}+b{x}^{2}+d$   

Also we have $P\left(−1\right)<P\left(1\right)$

$\Rightarrow\,1-a+b+d<1+a+b+d$

$\Rightarrow\,A>0$

Since ${P}^{\prime}\left(x\right)=0,$ only when $x=0$

and $P\left(x\right)$ is differentiable in $\left(−1,1\right)$, we should have the maximum and minimum at the points

$x=−1,0$ and $1$ only.

Also, we have $P\left(−1\right)<P\left(1\right)$

So,Maximum of $P\left(x\right)=Max\left\{P\left(0\right),P\left(1\right)\right\}$ and
Minimum of $P\left(x\right)=Min\left\{P\left(−1\right),P\left(0\right)\right\}$

In the interval $\left[0,1\right]$

${P}^{\prime}{\left(x\right)}=4{x}^{3}+3a{x}^{2}+2bx=x\left(4{x}^{2}+3ax+2b\right)$

Since ${P}^{\prime}{\left(x\right)}$ has only one root $x=0$, then $4{x}^{2}+3ax+2b=0$ has no real roots.

So,${\left(3a\right)}^{2}-32b<0$

$\Rightarrow\,\dfrac{3{a}^{2}}{32}>b$

So,$b>0$

Thus, we have $a>0$ and $b>0$

So,${P}^{\prime}{\left(x\right)}=4{x}^{3}+3a{x}^{2}+2bx>0,$ for $x\in\left(0,1\right)$

Hence, $P\left(x\right)$ is increasing in $\left[0,1\right]$ and $P\left(x\right)$ is decreasing in $\left[−1,0\right]$

Therefore, Maximum of $P\left(x\right)=P\left(1\right)$ and Minimum $P\left(x\right)$ does not occur at $x=−1$ respectively.

If $o<\alpha<\beta<\gamma<\dfrac {\pi}{2}$, then the equation $\dfrac {1}{x-\sin \alpha}+\dfrac {1}{x-\sin\beta}+\dfrac {1}{x-\sin \gamma}=0$ has

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

  1. Imaginary roots

  2. Real and equal roots

  3. Real and unequal roots

  4. Rational roots

Show answer
Correct Option: A
Explanation:

$\begin{array}{l} \frac { 1 }{ { x-\sin  \alpha  } } +\frac { 1 }{ { x-\sin  \beta  } } +\frac { 1 }{ { x-\sin  \gamma  } } =0 \ 0<\alpha <\beta <\gamma <\frac { \pi  }{ 2 }  \ let\, \alpha ={ 30^{ 0 } },\beta ={ 45^{ 0 } },\gamma ={ 60^{ 0 } } \ \Rightarrow \frac { 1 }{ { x-\frac { 1 }{ 2 }  } } =\frac { 1 }{ { x-2\sqrt { 2 }  } } =\frac { 1 }{ { x-\sqrt { \frac { 3 }{ 2 }  }  } } =0 \end{array}$

hence roots are real and unequal

The polynomial $\displaystyle (ax^{2}+bx+c)(ax^{2}-dx-c),ac\neq 0,$ has

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

  1. four real zeros

  2. at least two real zeros

  3. at most two real zeros

  4. no real zeros

Show answer
Correct Option: B
Explanation:

For 
$ax^{2}+bx+c=0$
$b^{2}-4ac\geq 0$ for real roots ...(i)
and for 
$ax^{2}-dx-c=0$
$d^{2}+4ac\geq 0$ for real roots ...(ii)
Now, 
$ac\neq 0$
Hence, 
Case I
If $ac>0$
Hence, 
$ax^{2}-dx-x=0$ has positive roots.
Case II
If $ac<0$
Then,
$ax^{2}+bx+c=0$
has Real roots.
Hence, the above polynomial has atleast two real roots.

If $\alpha$ and $\beta$ are the zeros of polynomial $x^{2}-ax+b$, then the value of $\alpha^{2}\left(\dfrac {\alpha^{2}}{\beta}-\beta\right)+\beta^{2}\left(\dfrac {\beta^{2}}{\alpha}-\alpha\right)$ is

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

  1. $\dfrac {a(a^{2}-4b)(a^{2}-b)}{b}$

  2. $\dfrac {b(a^{2}-4b)(a^{2}-b)}{a}$

  3. $\dfrac {b^{2}(a^{2}-4b)(a^{2}-b)}{a}$

  4. None 

Show answer
Correct Option: A
Explanation:
We have Sum of the roots$=\alpha+\beta=a$
Product of the roots$=\alpha\beta=b$

${\alpha}^{2}\left(\dfrac{{\alpha}^{2}}{\beta}-\beta\right)+{\beta}^{2}\left(\dfrac{{\beta}^{2}}{\alpha}-\alpha\right)$
$=\dfrac{{\alpha}^{2}}{\beta}\left({\alpha}^{2}-{\beta}^{2}\right)+\dfrac{{\beta}^{2}}{\alpha}\left({\beta}^{2}-{\alpha}^{2}\right)$
$=\dfrac{{\alpha}^{2}}{\beta}\left({\alpha}^{2}-{\beta}^{2}\right)-\dfrac{{\beta}^{2}}{\alpha}\left({\alpha}^{2}-{\beta}^{2}\right)$
$=\left({\alpha}^{2}-{\beta}^{2}\right)\left(\dfrac{{\alpha}^{2}}{\beta}-\dfrac{{\beta}^{2}}{\alpha}\right)$
$=\dfrac{\left({\alpha}^{2}-{\beta}^{2}\right)}{\alpha\beta}\left({\alpha}^{3}-{\beta}^{3}\right)$
$=\dfrac{\left(\alpha-\beta\right)\left(\alpha+\beta\right)}{\alpha\beta}\left(\alpha-\beta\right)\left({\alpha}^{2}+{\beta}^{2}+\alpha\beta\right)$
$=\dfrac{{\left(\alpha-\beta\right)}^{2}\left(\alpha+\beta\right)}{\alpha\beta}\left({\alpha}^{2}+{\beta}^{2}+\alpha\beta\right)$

We know that ${\alpha}^{2}+{\beta}^{2}={\left(\alpha+\beta\right)}^{2}-2\alpha\beta$ and 
${\left(\alpha-\beta\right)}^{2}={\left(\alpha+\beta\right)}^{2}-4\alpha\beta$

Using $\alpha+\beta=a$ and $\alpha\beta=b$ we have
${\left(\alpha-\beta\right)}^{2}={\left(\alpha-\beta\right)}^{2}={a}^{2}-4b$
And ${\alpha}^{2}+{\beta}^{2}+\alpha\beta={\left(\alpha+\beta\right)}^{2}-2\alpha\beta+\alpha\beta$
$={\left(\alpha+\beta\right)}^{2}-\alpha\beta={a}^{2}-b$
$=\dfrac{a\left({a}^{2}-4b\right)\left({a}^{2}-b\right)}{b}$



The value of $'a'$ for which the equation ${ x }^{ 3 }+ax+1=0$ and ${ x }^{ 4 }+a{ x }^{ 2 }+1=0$, have a common root is

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

  1. $a=2$

  2. $a=-2$

  3. $a=0$

  4. None of these

Show answer
Correct Option: B
Explanation:

Consider the following equation
$x^{4}+ax^{2}+1=0$
$x^{3}+ax+1=0$
Subtracting equation (ii) from (i), we get 
$x^{4}-x^{3}+a(x^{2}-x)=0$
$x^{3}(x-1)+ax(x-1)=0$
$(x-1)(x^{3}+ax)=0$
$x(x-1)(x^{2}+a)=0$
Hence, we get $x=0$ $x=1$ and $x^{2}=-a$
Now out of the above two, $x=0$ is not a root of the following two equations.
We do not know the nature of '$a$'. 

Hence, we cannot determine that $x^{2}=-a$ will have real or imaginary roots.
Hence, we get $x=1$ as a common root for the above two equations.
Now for both the equations to have $x=1$ as a common root, 
$f(1)=0$
$1+a+1=0$
$a=-2$
Similarly substituting in the second equation, we get $a=-2$.
Hence, the required value of $a$ is $-2$.

Coordinates of a point P are $(a, b)$ where $a$ is a root of the equation 

$x^{2}+x-42=0$ 
and $b$ is an integral root of the equation
$x^{2}+ax+a^{2}-37=0$. 
The coordinates of P can be

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

  1. $(6, 4)$

  2. $(-7, 4)$

  3. $(-7, 3)$

  4. $(6, -3)$

Show answer
Correct Option: B,C
Explanation:

$a^{2}+a-42=0\Rightarrow a=-7$ or $a=6.$
Since b is a root of $x^{2}+ax+a^{2}-37=0$
For $a=-7,$ we have $x^{2}-7x+49-37=0$
$\Rightarrow x^{2}-7x+12=0\Rightarrow x=4, 3 , so, b=4$ or $3.$
So the coordinates of P can be $(-7, 4)$ or $(-7, 3)$, 

For $a=6,$ we have $x^{2}+6x-1=0$ which does not give an integral value, so $a\neq 6.$

lf the difference of the roots of the equation $x^{2}-bx+c=0$ is equal to the differecne of the roots of the equation ${x}^{2}-{c}x+b=0$ and $b\neq c$, then $b+c=$

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

  1. $ 0$

  2. $2$

  3. $4$

  4. $-4$

Show answer
Correct Option: D
Explanation:

Let $(\alpha, \beta )$ and $ (\gamma, \delta )$ be the roots of the first equation and second equation respectively. 

Then, for the first equation
$ \alpha +\beta =b$ and $ \alpha \beta =c$
Now $ (\alpha +\beta )^{ 2 }={ b }^{ 2 }$
$ \Rightarrow (\alpha -\beta )^{ 2 }+4\alpha \beta ={ b }^{ 2 }$
$ \Rightarrow (\alpha -\beta )^{ 2 }={ b }^{ 2 }-4\alpha \beta $
$ \Rightarrow |\alpha -\beta |=\sqrt { { b }^{ 2 }-4c } $
Similarly for the second equation
$ |\gamma -\delta |=\sqrt { { c }^{ 2 }-4b } $
As per the given condition,
$ \sqrt { { b }^{ 2 }-4c } =\sqrt { { c }^{ 2 }-4b } $
$\Rightarrow { b }^{ 2 }-4c={ c }^{ 2 }-4b$
$\Rightarrow { b }^{ 2 }-{ c }^{ 2 }=-4(b-c)$
$ \Rightarrow (b+c)(b-c)=-4(b-c)$
Therefore, option D is correct.

Let $\displaystyle a _{1}, a _{2},a _{3},a _{4},a _{5} \, \varepsilon \, R$ denote a rearrangement of equation $\displaystyle p _{1}x^{5}+p _{2}x^{3}+p _{3}x^{2}+p _{4}x+p _{5}=0$ then, equation  $\displaystyle a _{1}x^{4}+a _{2}x^{3}+a _{3}x^{2}+a _{4}x +a _{5}=0$ has 

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

  1. at least two real roots

  2. all four real roots

  3. only imaginary roots

  4. none of these

Show answer
Correct Option: A
Explanation:
$a _{1}x^{4}+a _{2}x^{3}+a _{3}x^{2}+a _{4}x+a _{5}=0$
for $ x=1 $
$a _{1}+a _{2}+a _{3}+a _{4}+a _{5}=0$
for the given set of equation,
sum of  $p _{1}+p _{2}+p _{3}+p _{4}+p _{5}=0$ 
So, $a _{1}+a _{2}+a _{3}+a _{4}+a _{5}\epsilon R $  for any set of arrangement
Hence at least two real roots.

The sum of the solutions of the equation $64(81^{x})-84(144^{x})+27(256^{x})=0$  is:

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

  1. $1$

  2. $\dfrac{3}{2}$

  3. $\dfrac{5}{2}$

  4. None of these

Show answer
Correct Option: B
Explanation:

Given, $64(81^{ x })-84(144^{ x })+27(256^{ x })=0$

$ \Rightarrow 64\left( \left( \cfrac { 9 }{ 16 }  \right) ^{ x } \right) ^{ 2 }-84\left( \cfrac { 9 }{ 16 }  \right) ^{ x }+27=0$
Let $\left( \cfrac { 9 }{ 16 }  \right) ^{ x }=t$
$64t^{ 2 }-84t+27=0$
Solving this, we get
$\Rightarrow t=\cfrac { 3 }{ 4 } ,t=\cfrac { 9 }{ 16 } \Rightarrow x=\cfrac { 1 }{ 2 } ,x=1$
Hence, sum of roots is 
$\cfrac { 1 }{ 2 } +1=\cfrac { 3 }{ 2 } $
Hence, option 'B' is correct.

If the sum of two roots of the equation $x^{4}+px^{3}+qx^{2}+rx+8=0$ is equal to the sum of the other two, then $p^{3}+8r=$

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

  1. $p^2 - 4pq$

  2. $2pq$

  3. $p^2 - pq$

  4. $4pq$

Show answer
Correct Option: D
Explanation:

Let the roots of the equation be $a,b,c,d$. 
As per the question,
$a+b = c+d$ 
From the theory of polynomials, 
$a+b+c+d = -p$
$ \Rightarrow a+b=c+d= \displaystyle \frac{-p}{2} $

Also,
$ab+ac+ad+bd+bc+cd  = q $
$ \Rightarrow (a+b)(c+d) +ab+cd  =q $
$ \Rightarrow ab+cd = q - \displaystyle \frac{p^2}{4} $

Also, 
$abc+abd+bcd+adc = -r $
$ \Rightarrow ab(c+d) +cd(a+b) = -r $
$ \Rightarrow \displaystyle \frac {-p}{2} (ab+cd) = -r $
$ \Rightarrow \displaystyle \frac {-p}{2} ( q - \displaystyle \frac{p^2}{4} ) = -r $
$ \Rightarrow -4pq + p^3 = -8r $
$ \Rightarrow p^3 + 8r = 4pq $