Tag: sign of quadratic expression

Questions Related to sign of quadratic expression

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

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

  1. $3$

  2. $5$

  3. $7$

  4. $1$

Show answer
Correct Option: D
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 .

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

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

  1. $ab+b^2+2ac=0$

  2. $ab+a^2+2ac=0$

  3. $ab+{b}^{2}-2ac=0$

  4. $ab+{a}^{2}-2ac=0$

Show answer
Correct Option: C
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 } $

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

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

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

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

  1. $ \pm 2$

  2. $\pm 3$

  3. $ 5$

  4. $-5$

Show answer
Correct Option: A
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 $
$\Rightarrow { \left( -p \right)  }^{ 2 }=10+2\left( -3 \right) =10-6=4$
$\Rightarrow { p }^{ 2 }=4$
$\Rightarrow p=\pm 2$

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

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

  1. $\displaystyle \frac{7}{8}$

  2. $-\displaystyle \frac{7}{8}$

  3. $\displaystyle \frac{9}{8}$

  4. $-\displaystyle \frac{9}{8}$

Show answer
Correct Option: B
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 } $

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

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

  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 $32$

Show answer
Correct Option: A,C,D

Find the equation $x^4+4rx+3s=0$ =0 has no real root, then 

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

  1. $r^2$

  2. $r^2>s^2$

  3. $r^4$

  4. $r^4>s^3$

Show answer
Correct Option: A

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

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

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

Show answer
Correct Option: A
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$

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:

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

  1. $4$

  2. $5$

  3. $6$

  4. $7$

Show answer
Correct Option: C
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 $6$

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

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

  1. pr = q

  2. qr = p

  3. pq = r

  4. $p ^ { 2 } q ^ { 2 } = r$

Show answer
Correct Option: C
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

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:

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

  1. $-4$

  2. $2$

  3. $6$

  4. cannot be determined

Show answer
Correct Option: B
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)
$\dfrac{1}{4} (\alpha + \beta + \gamma + \sigma) = 1$
$\Rightarrow \dfrac{1}{4} (\alpha + \beta + \gamma + \sigma) = (\alpha \beta \gamma \sigma) \dfrac{1}{4}$
$\therefore A. M. = a. m.$
$\therefore \alpha = \beta = \gamma = \sigma$
$4 \alpha = 4$
$\therefore \alpha = 1$
$1 - 4 + a + b + 1 = 0$
$\therefore a + b = 2$