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 is the Descartes rule used to find the number of roots in an 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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

  1. four real and roots

  2. two irrational roots

  3. one integer roots

  4. two imaginary roots
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation is a reciprocal equation of the first kind. Dividing by x^2 and substituting y = x + 1/x, the equation becomes 12(y^2 - 2) - 56y + 89 = 0, or 12y^2 - 56y + 65 = 0. Solving for y gives y = 5/2 and y = 13/6. Solving x + 1/x = y for these values yields four real roots.

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

  1. Can be imaginary or real

  2. Must be real

  3. Must be Imaginary

  4. must be zero
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  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 $(-\pi)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let f(x) = x^3 - (pi-1)x^2 - pi. Testing values: f(1) = 1 - pi + 1 - pi = 2 - 2pi < 0. f(pi) = pi^3 - (pi-1)pi^2 - pi = pi^3 - pi^3 + pi^2 - pi = pi^2 - pi > 0. There is a real root between 1 and pi. Further analysis shows the other two roots are 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

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 $(s)$.

  1. $(-2,\,0)$
  2. $[0,\,1]$
  3. $[1,\,2]$
  4. $\left( { - \infty ,\,\infty } \right)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a cubic to have two distinct real roots in [0, 1], the function must have a local maximum and minimum within or near the interval, and the values at the endpoints must satisfy specific conditions. Analysis of the derivative 9x^2 + 2x - 7 = 0 gives roots at x = -1 and x = 7/9. Testing these in the cubic leads to the interval [0, 1].

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 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 

  1. At least one root

  2. No root

  3. Exactly one root

  4. At most two roots
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let f(x) = 2x^3 + 4x^2 - 4x - 2. By Rolle's Theorem, if f(alpha) = f(beta) = 0, then f'(x) = 6x^2 + 8x - 4 has a root between alpha and beta. The second equation is g(x) = 8x^3 + 3x^2 - 2. Analysis of the derivatives and signs shows exactly one root exists between the roots of f(x).

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

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

  1. $-1$
  2. $1$
  3. $2$
  4. $3$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $a=1$

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

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?

  1. $330$
  2. $338$
  3. $350$
  4. $360$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation is x^3 + (112-2k)x^2 + 112x - 1 = 0. Given roots alpha, beta are positive integers with beta < 4. Testing integer values for beta (1, 2, 3) and using Vieta's formulas, we find alpha = 330, beta = 1, etc. The sum alpha + beta + alpha*beta is consistent with 330.

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

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
$0 < 3ab \le 1$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
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, $0<3ab\le1.$