Tag: algebraic functions, equations and inequalities

Questions Related to algebraic functions, equations and inequalities

How is the Descartes rule used to find the number of roots in an equation?

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

  1. By counting the number of times the equation changes signs

  2. By counting positive signs in the equation

  3. By counting negative signs in the equation

  4. None of the above

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Correct Option: A
Explanation:

Descartes rule counts the number of times the sign changes from either $+$ to $-$ or from $-$ to $+$

Equation $12x^4-56x^3+89x^2-56x+12=0$ 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 and roots

  2. two irrational roots

  3. one integer roots

  4. two imaginary roots

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Correct Option: A

If one root of a cubic equation is real and second root is imaginary, then what can be said about the third root?

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

  1. Can be imaginary or real

  2. Must be real

  3. Must be Imaginary

  4. must be zero

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Correct Option: C
Explanation:

Cubic Equations can have at max $3$ roots. Also, imaginary roots always occur in a pair of conjugates.
Since here one root is real and the other is imaginary, the third one must be imaginary and it will be the conjugate of the second root.

The equation $x^3 + 6x^2 + 11x + 6 = 0$ has

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

  1. no negative roots

  2. no positive real roots

  3. no real roots

  4. $1$ positive and $2$ negative roots

  5. $1$ negative and $2$ positive roots

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Correct Option: B
Explanation:

$x^3+6x^2+11x+6=0$

$(x+1)(x+2)(+3)=0$
$x=-1,-2,-3$
So, no positive real roots.

The roots of the cubic $x^{3} - (\pi - 1)x^{2} - \pi = 0$, are

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

  1. All three real and distinct

  2. One real and two coincident

  3. One real and two imaginary with product of the imaginary roots being $\pi$

  4. One real, two imaginary with sum of the imaginary roots being $(-\pi)$

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Correct Option: C

The real value of $\lambda $ for which the equation, $3{x^3} + {x^2} - 7x + \lambda  = 0$, has two distinct real roots in $[0,\,1]$ lie in the interval $(s)$.

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

  1. $(-2,\,0)$

  2. $[0,\,1]$

  3. $[1,\,2]$

  4. $\left( { - \infty ,\,\infty } \right)$

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Correct Option: B

lf the equation $4 x ^ { 2 } + 2 x ^ { 3 }-4 x - 2 = 0$ has two real roots $\alpha \text { and } \beta$ then between $\alpha \text { and } \beta$ the equation $8 x ^ { 3 } + 3 x ^ { 2 } - 2 = 0$ 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 one root

  2. No root

  3. Exactly one root

  4. At most two roots

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Correct Option: C

The values for  which ${x^4} - 2a{x^2} + {a^2} - a = 0$ has all real 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. $-1$

  2. $1$

  3. $2$

  4. $3$

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Correct Option: B
Explanation:

${x}^{4}-2a{x}^{2}+{a}^{2}-a=0$

This equation can also be written in quadratic form as
${({x}^{2})}^{2}=2a({x}^{2})+{a}^{2}-a=0$
Substituting ${x}^{2}=t$ where $t> 0$
we get
${t}^{2}-2Aat+{a}^{2}-a=0$
Now, for quadratic equation to have real rpots
${(-2a)}^{2}-4\times 1\times ({a}^{2}-a)\ge 0$
(...applied the condition for real roots of quadratic equation $a{x}^{2}+bx+c=0$ ${b}^{2}-4ac\ge 0$)
Hence
$4{A}^{2}-4({a}^{2}-a)\ge 0$
$\Rightarrow$ $4{a}^{2}-4{a}^{2}+4a\ge 0$
$4a\ge 0$
$a\ge 0$
Also we have $t\ge 0$
$\cfrac { -(2a)\pm \sqrt { 4a }  }{ 2.1 } \ge \quad 0$
$\cfrac { -(2a)\pm 2\sqrt { a }  }{ 2.1 } \ge \quad 0$
$-a\pm \sqrt { a } \ge 0$
This gives us the only solution possible from options given as $a=1$

Consider the equation $x^3+(112-2k)x^2+110x+2x-1=0$ having two positive integral roots $\alpha$ and $\beta$(where $\beta < 4, k\in R)$.
The value of $\alpha +\beta +\alpha\beta$ is?

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

  1. $330$

  2. $338$

  3. $350$

  4. $360$

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Correct Option: A

Suppose $a$ and $b$ are real no. such that the roots of the cubic equation $ax^{3}-x^{2}+bx+1=0$ are all positive real no. then
$0 < 3ab \le 1$

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

  1. True

  2. False

Show answer
Correct Option: A
Explanation:
The given equation is $ax^3-x^2+bx+1=0$
Let $\alpha,\,\beta,\,\gamma$ be the roots of the given equation.
We have
$\alpha+\beta+\gamma=\dfrac{1}{a}$
$\alpha\beta+\beta\gamma+\gamma\alpha=\dfrac{b}{a}$
$\alpha\beta\gamma=\dfrac{1}{a}$
It follows that $a,b$ are positive. we obtain
$\dfrac{3b}{a}=3(\alpha\beta+\beta\gamma+\gamma\alpha)\le(\alpha+\beta+\gamma)^2=\dfrac{1}{a^2}$
Which gives, $0<3ab\le1.$