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Multiple choice reciprocal equations theory of equations maths

The root(s) of the reciprocal equation of second type and of even degree is/are

  1. $x=1$
  2. $x=-1$
  3. $x=\pm1$
  4. $x=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$x = -1$ is a root of the reciprocal equation of first type and of odd degree. 
$x = 1$ is a root of the reciprocal equation of second type and of odd degree.
$x=±1$ are two roots of reciprocal equation of second type and of even degree. 
Multiple choice reciprocal equations theory of equations maths

lf $\mathrm{f}({x})=0$ is a reciprocal equation of second type and even degree, then a factor of $\mathrm{f}({x})$  is:

  1. $x+1$
  2. $x-1$
  3. $x^{2}-1$
  4. $x^{2}$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

The equation will have 2 solutions, 1 and -1.
The solution will be x+1 and x-1,
$(x-1)(x+1)=x^2-1$

Multiple choice reciprocal equations theory of equations maths

The equation whose roots are the reciprocal of the roots of $2x^2 - 3x -5=0$, is:

  1. $5x^2+3x-2=0$
  2. $2x^2+3x-5=0$
  3. $3x^2-3x+2=0$
  4. $2x^2+5x -3 = 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$2x^2-3x-5=0$


$\Rightarrow (x+1)(2x-5)=0$

$\therefore \alpha=-1$ and $\beta=\dfrac{5}{2}$

Reciprocal of these roots, $\alpha=-1$ and $\beta=\dfrac{2}{5}$

General form of quadratic equation $x^2-(\alpha+\beta)x+\alpha\beta=0$

$\Rightarrow x^2-(-1+\dfrac{2}{5})x-\dfrac{2}{5}=0$

$\Rightarrow 5x^2+3x-2=0$

Multiple choice reciprocal equations theory of equations maths

The root of the reciprocal equation of first type and of odd degree is:

  1. $x= 1$
  2. $x=-1$
  3. $x=\pm1$
  4. $x=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$x = -1$ is a root of the reciprocal equation of first type and of odd degree. 
$x = 1$ is a root of the reciprocal equation of second type and of odd degree.
$x=±1$ are two roots of reciprocal equation of second type and of even degree. 
Multiple choice reciprocal equations theory of equations maths

If the reciprocal of every root of an equation is also a root of it, then the equation is said to be a

  1. reciprocal equation of first type

  2. reciprocal equation of second type

  3. reciprocal equation

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

An equation whose roots can be divided into pairs of numbers, each the reciprocal of the other or an equation which is unchanged if the variable is replaced by its reciprocal is known as reciprocal euation.

Multiple choice reciprocal equations theory of equations maths

lf $\mathrm{f}(\mathrm{x})=0$ is a reciprocal equation of first type and odd degree, then a factor of $\mathrm{f}(\mathrm{x})$ is:

  1. $\mathrm{x}-2$
  2. $\mathrm{x}-1$
  3. $\mathrm{x}$
  4. $\mathrm{x}+1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When the reciprocal equation is of an odd degree, $x=-1$ is always a solution.
So, $x+1$ is a factor.

Multiple choice reciprocal equations theory of equations maths

The root of the reciprocal equation of second type and of odd degree is:

  1. $x=-1$
  2. $x=+1$
  3. $x=\pm1$
  4. $x=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$x = -1$ is a root of the reciprocal equation of first type and of odd degree. 
$x = 1$ is a root of the reciprocal equation of second type and of odd degree.
$x=±1$ are two roots of reciprocal equation of second type and of even degree. 
Multiple choice reciprocal equations theory of equations maths

lf $\mathrm{f}({x})=0$ is a reciprocal equation of second type and fifth degree, then a root of $\mathrm{f}({x})=0$  is:

  1. $0$
  2. $1$
  3. $-1$
  4. $2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When the reciprocal equation is of an odd degree, second type, $x=1$ is always a solution.

Multiple choice reciprocal equations theory of equations maths

The roots equation $x^4-3x^3+4x^2-3x+1=0$ is 

  1. $0$
  2. $1$
  3. $2$
  4. $3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$x^4-3x^3+4x^2-3x+1=0$
This equation is resiprocal equation of first type as $a _{n-i}=a _i$
Dividing equation by $x^2$:
$x^2-3x+4-\dfrac3x+\dfrac{1}{x^2}=0$
$x^2+\dfrac{1}{x^2}-3x-\dfrac3x+4=0$
$\left(x+\dfrac1x\right)^2-2-3\left(x+\dfrac1x\right)+4=0$
$\left(x+\dfrac1x\right)^2-3\left(x+\dfrac1x\right)+2=0$
Let $x+\dfrac1x=y$
$y^2-3y+2=0$
$(y-1)(y-2)=0$
$y=1$ or $y=2$
For $y=1$:
 $x+\dfrac1x=1$
$x^2-x+1=0$
$D=(-1)^2-4(1)(1)=-3<0$
Hence no real value of x exists for this case.
For $y=2$:
$x+\dfrac1x=2$
$x^2-2x+1=0$
$(x-1)^2=0$
$x=1$
Hence solution of the given equation is x=1.

Multiple choice reciprocal equations theory of equations maths

If the coefficients from one end of an equation are equal in magnitude and sign to the coefficients from the other end, then the equation is said to be 

  1. reciprocal equation of second type

  2. reciprocal equation of first type

  3. reciprocal equation

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation