Tag: descartes rule

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

How many real solutions does the equation $x^{7}+14x^{5}+16x^{3}+30x-560=0$ has?

  1. $3$
  2. $5$
  3. $7$
  4. $1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let  $f(x) =x^{7}+14x^{5}+16x^{3}+30\mathrm{x} -560$

$\Rightarrow f'(x)=7x^{6}+70x^{4}+48x^{2}+30>0$  $\forall x \in R$

$\therefore f$ is increasing also $\displaystyle
\lim _{x\rightarrow\infty}f(x)=\infty$ ; $\displaystyle
\lim _{x\to-\infty}f(x)=-\infty$

Hence, $f(x) =0$ has exactly one real root .

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

lf the sum of the roots of the equation $ax^2+bx+c=0$ is equal to sum of their squares, then

  1. $ab+b^2+2ac=0$
  2. $ab+a^2+2ac=0$
  3. $ab+{b}^{2}-2ac=0$
  4. $ab+{a}^{2}-2ac=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $\alpha, \beta$ are roots of $\displaystyle a{ x }^{ 2 }+bx+c=0$, then


$\displaystyle \alpha +\beta =-\frac { b }{ a } $

$\displaystyle \alpha \beta =\frac { c }{ a } $

As sum of roots is equal to sum of their square, then 
$\displaystyle \alpha +\beta ={ \alpha  }^{ 2 }+{ \beta  }^{ 2 }$

$\displaystyle \Rightarrow \alpha +\beta ={ \left( \alpha +\beta  \right)  }^{ 2 }-2\alpha \beta $

$\displaystyle \Rightarrow -\frac { b }{ a } ={ \left( -\frac { b }{ a }  \right)  }^{ 2 }-2\left( \frac { c }{ a }  \right) $

$\displaystyle \Rightarrow -\frac { b }{ a } =\frac { { b }^{ 2 } }{ { a }^{ 2 } } -\frac { 2c }{ a } $

$\displaystyle \Rightarrow -ab={ b }^{ 2 }-2ac$

$\displaystyle \Rightarrow { b }^{ 2 }+ab-2ac=0$ 

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

lf the sum of the squares of the roots of $x^{2}+px-3=0$ is $10$, then $p=$

  1. $ \pm 2$
  2. $\pm 3$
  3. $ 5$
  4. $-5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $\alpha,\beta$ are roots of ${x}^{2}+px-3=0,$ then

${ S } _{ 1 }=\alpha +\beta =-p$
And ${ S } _{ 2 }=\alpha \beta =-3$

Given ${ \alpha  }^{ 2 }+{ \beta  }^{ 2 }=10$
Now from ${ \left( \alpha +\beta  \right)  }^{ 2 }={ \alpha  }^{ 2 }+{ \beta  }^{ 2 }+2\alpha \beta $
$\Rightarrow { \left( -p \right)  }^{ 2 }=10+2\left( -3 \right) =10-6=4$
$\Rightarrow { p }^{ 2 }=4$
$\Rightarrow p=\pm 2$

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

If the sum of two roots of the equation  $x^{4}-x^{3}+2x^{2}+kx+17=0$ equals to the sum of the other two, then $k $ is equal to

  1. $\displaystyle \frac{7}{8}$
  2. $-\displaystyle \frac{7}{8}$
  3. $\displaystyle \frac{9}{8}$
  4. $-\displaystyle \frac{9}{8}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $\alpha ,\beta ,\gamma ,\delta $ be the roots of ${ x }^{ 4 }-{ x }^{ 3 }+2{ x }^{ 2 }+kx+17=0$
Such that $\alpha +\beta =\gamma +\delta $
Then ${ s } _{ 1 }=\alpha +\beta +\gamma +\delta =1\Rightarrow \alpha +\beta =\cfrac { 1 }{ 2 } $ 
${ s } _{ 2 }=\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta =2\Rightarrow { \left( \alpha +\beta  \right)  }^{ 2 }+\alpha \beta +\gamma \delta =2$
$\Rightarrow \alpha \beta +\gamma \delta =2-\cfrac { 1 }{ 4 } =\cfrac { 7 }{ 4 } $   ...(1)
${ s } _{ 3 }=\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta =-k\Rightarrow \left( \alpha +\beta  \right) \left( \alpha \beta +\gamma \delta  \right) =-k$
$\Rightarrow \left( \alpha \beta +\gamma \delta  \right) =-2k$   ...(2)
From (1) and (2), we have
$-2k=\cfrac { 7 }{ 4 } \Rightarrow k=-\cfrac { 7 }{ 8 } $

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

Let $P(x) = x^{32} - x^{25} + x^{18} - x^{11} + x^{4} - x^{3} + 1$. Which of the following are CORRECT?

  1. Number of real roots of $P(x) = 0$ are zero
  2. Number of imaginary roots of $P(x) = 0$ are $32$
  3. Number of negative roots of $P(x) = 0$ are zero
  4. Number of imaginary roots of $P(x) + P(-x) = 0$ are $32$
Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
Explanation

P(x) = x^32 - x^25 + x^18 - x^11 + x^4 - x^3 + 1. For x >= 1, P(x) > 0. For 0 <= x < 1, P(x) > 0. For x < 0, let x = -y (y > 0), P(-y) = y^32 + y^25 + y^18 + y^11 + y^4 + y^3 + 1 > 0. Thus, there are no real roots. Since there are no real roots, all 32 roots must be imaginary. P(-x) also has no real roots, so its roots are also imaginary.

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

If the sum of two of the roots of $x^4-2x^3-3x^2+10x-10=0$ is zero then the roots are

  1. $\pm \sqrt{5},1\pm i$
  2. $\pm \sqrt{5},1-i$
  3. $\large{\frac{1}{2}},-\large{\frac{1}{5}},\pm 1$
  4. $\sqrt{2},\sqrt{5},\pm 2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

let roots are $\pm a,b,c\ b+c=2\ -{ a }^{ 2 }bc=-10\ { a }^{ 2 }bc=10\ { -a }^{ 2 }+ab+ac+bc-ab-ac=-3\ bc-{ a }^{ 2 }=-3\ { a }^{ 2 }-bc=3\ $

let $bc=t$
from $(2) t=\frac { 10 }{ { a }^{ 2 } } \ (3)\quad \quad { a }^{ 2 }-t=3\ { a }^{ 2 }-\frac { 10 }{ { a }^{ 2 } } =3\ { a }^{ 4 }-{ 3a }^{ 2 }-10=0\ { a }^{ 4 }-{ 5a }^{ 2 }+{ 2a }^{ 2 }-10=0\ { a }^{ 2 }\left( { a }^{ 2 }-5 \right) +2\left( { a }^{ 2 }-5 \right) =0\ \left( { a }^{ 2 }-5 \right) \left( { a }^{ 2 }+2 \right) =0\ a=\pm \sqrt { 5 } \ bc=2\ c=\cfrac { 2 }{ b } \ b+\cfrac { 2 }{ b } =2\ { b }^{ 2 }-2b+2=0$
$\quad \quad b = 1 \pm i$
$ \quad \quad c = \cfrac{2}{b} = \cfrac{2}{1\pm i} = 1 \mp i$
$ \therefore $ roots are $ 1\pm i, \pm\sqrt5$

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

A polynomial of 6th degree $f(x)$ satisfies $f(x)=f(2-x),:\forall:x\epsilon R$, if $f(x)=0$ has 4 distinct and two equal roots, then sum of the roots of $f(x)=0$ is:

  1. $4$
  2. $5$
  3. $6$
  4. $7$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In the functional relation replace $x$ with $x+1$.

We have,
$f(1+x)=f(1-x)$
This shows that the function is symmetric about $x=1$.
There is one and only one double root. If the double root exists at any value $x _0$ other than at $x=1$, then a double root will also exist at a value of $2-x _0$.
Hence, the double root exists at $x=1$ 
Say two other roots are $\alpha$ and $\beta$
$f(\alpha)=f(2-\alpha)=0$
$\therefore 2-\alpha$ is also a root.
And similarly, $2-\beta$ is also a root.
$\therefore$ the roots are $1, 1, \alpha, \beta, 2-\alpha, 2-\beta$
Hence, sum of the roots is $6$

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

If two roots of the equations $x ^ { 3 } - p x ^ { 2 } + q x - r = 0$ are equal in magnitude but opposite in sign, for

  1. pr = q

  2. qr = p

  3. pq = r
  4. $p ^ { 2 } q ^ { 2 } = r$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation 
let those are m, -m
now sum of three roots = p
hence third root will be p
now 
m*(-m) + m*p + (-m)*p = q
hence  –m2 = q
now m*( –m) * p = r
 –m2 p  = r
put value of  –m2 = q
hence  pq = r

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

If the equation ${x}^{4}-4{x}^{3}+a{x}^{2}+bx+1=0$ has four positive roots, then the value of $(a+b)$ is:

  1. $-4$
  2. $2$
  3. $6$

  4. cannot be determined
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $x^4 - 4x^3 + ax^2 + bx + 1 = 0$

let the root of equation be $\alpha, \beta, \gamma, \sigma$
$\alpha + \beta + \gamma + \sigma = 4$ ...(i)
$\alpha \beta \gamma \sigma = 1$ ... (ii)
$\dfrac{1}{4} (\alpha + \beta + \gamma + \sigma) = 1$
$\Rightarrow \dfrac{1}{4} (\alpha + \beta + \gamma + \sigma) = (\alpha \beta \gamma \sigma) \dfrac{1}{4}$
$\therefore A. M. = a. m.$
$\therefore \alpha = \beta = \gamma = \sigma$
$4 \alpha = 4$
$\therefore \alpha = 1$
$1 - 4 + a + b + 1 = 0$
$\therefore a + b = 2$